i,^:j Cornell University Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031237849 Cornell University Library 3 1924 031 237 849 ELEMENTS PLAIfE AID SPHEEICAL TEIGOKOIETEY; WITH PRACTICAL APPLICATIONSc By benjamin GEEENLEAF, A.M., AUTHOR OF A MATHEMATICAL SERIES. IMPROVED ELECTEOTYPE EDITION. /CORNELL university! DG ED . . , ED^ 0^ = sm«sm5, we have cos (a — ^) = cos a cos 5 -f- sin a sin 5. (20) 64. The four formulae last established apply to arcs as well as angles, and may be considered the fundamental formidce of sub- sequent analysis. They are brought together in the following TABLE. 1. sin {a-\-h) =: sin a cosh -\- cos a si]i b. 2. cos {a-\-V) = cos a cos 6 — sin a sin b. 3. sin (cr — b) = sin a cos 6 — cos a sin b. 4. cos (a — 6) = cos a cos b -\- sin a sin S. BOOK II. 23 A" A' SIGNS OF THE TRIGONOMETRIC FUNCTIONS. 65. If on any line, straight oi- curved, different distances be measured from a fixed point of origin, the distances which have contrary situations may by convention be introduced into our calculations, by affecting the quantities representing their mag- nitudes by contrary signs. Let be a fixed point in any line, A A", and suppose we have to determine the situations of other points in this line with respect to 0. The position of any point in the line will be known if we know the distance of the point from 0, and also know on which side of the point lies. Now it is found convenient to adopt the following convention: dis- tances measured in one direction from along the line will be denoted by positive quantities, and .distances measured in the contrary direc- tion from will be denoted by negative quantities. Thus, for example, suppose that distances measured from towards the right are denoted by positive numbers, and let A be a point, the distance of which from is denoted by 2 or -(- 2 ; then if .^4 ' be a point situated just as far to the left of as A is to the right, the distance of A" from will be denoted by —2. In like manner, if distances originating in A A", and taken along A', or only parallel to A', when measured upwards be denoted by positive quantities, on being measured downwards they will be denoted by negative quantities. 66. A similar convention may con- A veniently be adopted with respect to angles. Let any line, B, revolve from the position OA, in one direction round 0, forming the angle BOA, and let this angle be denoted by a positive quan- tity ; then, if the line B revolve, 24 TRIGONOMETRY. from the position A, round in the contrary direction, form- ing the angle BOA, this angle may be denoted by a negative quantity. Thus, for example, if each of the angles A OB and A B is two ninths of a right angle, and we denote the former by 20° or +20°, the latter may be denoted by — 20°. The direction of the positive distances is quite indiiferent ; but, being once fixed, the negative distances must lie in the contrary direction. 67. The representation of angles as the measure of the revo- lution of a line, turning in a plane about one of its own points, leads to the consideration of angles, not only greater than two right angles, but of all possible magnitudes. Thus, when the line B, starting from the initial position A, has passed A", or made more than half a revolution, we have described an angular magnitude of more than 180°; and when it has passed on to A, we have an angular magnitude of 360°. If it now contin- ues to revolve in the same direction till it arrives again at B, we have an an- gular magnitude of 360°+ 20°= 380°, and thus . we may conceive of angles of all magnitudes. In like manner negative angles of all magnitudes may be formed by the describ- ing line B revolving from A, but in a contrary direction. 68. The algebraic signs of the trigonometric functions can be readily fixed in the mind by being repissented geometrically. Thus, Let the extremity of a revolving line, starting from the initial posi- tion A, describe the positive arc A B less than 90°, A B between 90° and 180°, AA'B" between 180° and 270°, and AAA"B'' between 270° and 360°. Then, according to the definitions of Art. 54, BD, BD'i BOOK II. 25 Bf'D', and B'"D are the sines, and D and D' are the cosines, of the angles measured by the arcs terminating in each of the four quadrants. As all the functions of an angle less than 90° are considered positive, their direction will fix the signs for other quadrants. It will be seen (Art. 65) that the sines are above the diameter A A", and positive, in the first and second quadrants, but helow, and negative, in the third' and fourth quadrants. Also (Art. 65), the cosines are to the right of the diameter A' A!", and positive, in the first and fourth quadrants, but to the left, and negative, in the second and third quadrants. As tan ^ = - . (13), the tangent must be positive when the sine and cosine have the same sign, and negative when they have unlike signs. Hence the tangents are positive in the first and third quadrants, and negative in the second and fourth quadrants, a result which may also be obtained by noticing whether the tangents must run above or helow the origin A, to meet the secant. The cotangent of any angle or arc always has the same alge- braic sign as its tangent, the secant the same as the cosine, and the cosecant the same as the sine ; for they are reciprocals (Art. 51). The versed sine, coversed sine, and su versed sine, since they are referred to the' origins of their arcs, A, A', and A", as fixed points, instead of the centre 0, can have but one direction, and therefore are always positive. By comparing the sine B"'D and the cosine D oi the nega- tive arc A B" (Art. 66) with those of the equal positive arc A B, it will be seen that the cosines are identical, and conse- quently the secants are equal ; but the sines, and, consequentlj', the tangents, cotangents, and cosecants, have unlike algebraic signs. The functions of the arc A B", terminating in the first negative quadrant, are the same as those of the arc A A'A"B"', terminating in the fourth positive quadrant. The second nega- tive and third positive, the third negative and second positive, and the fourth negative and first positive quadrants likewise have functions with the same algebraic signs. 3* 26 TRIGONOMETRY. 69. From the results above obtained is formed the following TABLE. Positive Quadrant. Sine. Cosine. Tangent. Cotan't Secant. Cosecant. Negative Quadrant. First. + + + + + + Fourth. Second. + — — — — + Third. Third. — — + + — — Second. Fourth. — + — — + — First. Values of the Trigonometric Functions op certain Angles. 70. The definitions of the trigonometric functions already given (Art. 46-50) apply directly only to angles not exceeding a right angle. But by means of the formulae which have been deduced from them we may now extend the definitions so as to render them applicable to angles of any magnitude. 71. To find the sine, ^c. of 30° and of 60°. Let ABO be an equilateral triangle, then each of its angles equals one third of two right angles, or 60°. Draw BD perpendicular to A O, then the angle A BD is equal to ^ AB Q, AB is equal to Z) C, and AD^^^ B G= i AB. Therefore, sm A BB := -r-^ zz= 2 ^- 1 . A B AB *' or, since 30° and 60° are complements the one of the other, sin 30° = cos 60° = ^, (21) whence by (10) cos 30° = sin 60° = v/I^ = J- v^3. _ (22) Then, by (13) and (o), tan30° = cot60°=^-i^ = J^ = i,/3, (23) BOOK 11. 27 cot 30° = tan 60° = ^- = y/3, (24) sec 30° = cosec 00° = /, , = ^ = | v^S, (25) cosec 30° = sec 60° = ^ = 2. (26) 72. Tb fnd the sine, ^c. o/ 45°. Since 45° is the complement of 45°, sin 45° = cos 45°. Then making A = 45° in (8), we have sin^ 45° 4- cos^ 45° = 2 sin^ 45° = 2 cos* 45° = 1, or sjn^ 45° = cos* 45° = ^, sin 45° = cos 45° := y/^ = ^ y/2. (27) Hence by (13) and (5), ' tan 45° = cot 45° = f ^ = 1, (28) sec 45° = cosec 45° = j-^ = v/2. (29) 73. To find the sine, ^c. of 0° and of 90°. Since 0° and 90° are complements the one of the other, sin 0° = cos 90°. Then making a ^ S in (19) and (20), we have sin 0° = cos 90° ^ sin a cos a — cos a sin a = 0, (30) cos 0° = sin 90° = cos a co3 a -\- sin a sin a, or by (8), cos 0° = sin 90° = cos*a + sin*o = 1. (31) Hence by (13) and (5), tan 0° = cot 90° = a = 0, (32) cot 0° = tan 90° = J = 00 , (33) sec 0° = cosec 90° = \=l, (34) cosec 0° = sec 90° = ^ = oo. (35) 28 TRIGONOMETRY. 74. To find the sine, S^c. of 180°. Let a = b= 90° in (17) and (18) ; then, by means of (30) and (31), we have sin 180° = 1X0 + 0X1 = 0, (36) cos 180° = 0x0—1X1= — 1- (37) by (13) and (5), tan 180° = -^ = 0, cot 180° = i- = oo , (38) sec 180°=-^ = — 1, cosec 180° = i- = 00 . (39) 75. To find the sine, Sfc. of 270°. Let a = 180° and b = 90° in (17) and (18), and we have sin 270° = X + (— 1) X 1 = — 1, (40) cos 270° = (— 1) X — X 1 = 0. (41) Hence, by (13) and (5), tan 270° = =i = oo , cot 270° = -^ = 0, . (42) sec 270° = 1 = 00 , cosec 270° = ^ = —1. (43) 76. To find the sine, ^-c. of 360°. Let a = 6 = 180° in (17) and (18), and we have sin 360° = X (— 1) + (— 1) X = 0, (44) cos 360° = (— 1) X (— 1) — X = 1. (45) But these values for the sine and cosine of 360° are the same as those for tlie sine and cosine of 0°. Hence, All the trigonometric functions of 360° are the same as those for 0°. 77. To find the sine, S^c. of the supplement of an angle. Let a = 180° in (19) and (20) ; then, by means of (36) and (37), we have sin (180° — b) = sin b, cos (180° — 5) = —cos i (46) BOOK II. whence, by (13) and (5), tan (180° - ,. sin J - *) = — ; COS cot (180° - -h)- ' "1 — _tan b sec (180° - b) - ^ ^ — cos b cosec (180° - ' sin b 29 = —tan 5, (47) = —cot h, (48) = —sec 5, (49) = cosec h ; (50) that is, the sine and cosecant of the supplement of an angle are the same as those of the angle itself; and the cosine, tangent, cotangent, and secant are the negatives of those of the angle. 78. It follows from the preceding article, that the sine and cosecant of an obtuse angle are positive, while its cosine, tangent, cotangent, and secant are negative, as has before been shown geometrically (Art. 68, 69). 79. To find the sine, Src. of a negative angle. Let a = 0° in (19) and (20) ; then, by means of (30), (31), (13), and (5), we have sin ( — b) = — sin b, cos ( — b) =^ cos b, (51) tan ( — 5) = — tan b, cot ( — b) =^ — cot b, (52) sec ( — b) = sec b, cosec ( — b) = — cosec b; (53) that is, the cosine and secant of the negative of an angle are the same as those of the angle itself ; and the sine, tangent, cotangent, and cosecant of the negative of an angle are the negatives of those of the angle. These results correspond with those obtained geometrically (Art. 68). 80. To find the sine, ^c. of an angle which exceeds 180°. Let a = 180° in (17) and (18) ; then, by means of (36) and (37), we have sin (180°+ 6) =— sin J, cos (180° + J) = —cos S, (54) tan (180° + J) = tan b, cot (180° + 5) = cot b, (55) sec (180°+ 6) = —sec b, cosec (180° + J) = —cosec 5; (56) 30 TEIGONOJIETEV. that is, the tangent and cotangent of an angle which exceeds 180*^ are equal to those of its excess above 180°; and the sine, cosine, secant, and cosecant of this angle are the negatives of those of its excess. 81. It follows from the precedifig article, that the tangent and cotangent of an angle which exceeds 1 80° and is less than 270^ are positive; while its sine, cosine, secant, and cosecant are negative. Also, by considering h greater than 90° (Art. 78), that the cosine and secant of an angle which exceeds 270° and is less than 360° are positive ; while its sine, tangent, cotangent, and cosecant are negative. (See Art. 68, 69.) 82. To find the sine, S^c. of an angle which exceeds 360°. Let a = 360° in (17) and (18) ; then, by means of (44) and (45), we have sin (360° +h)— sin h, cos (360° + 5) = cos S ; (57) that is, all the trigonometric functions of an angle which exceeds 360° are the same as those of the excess above 360°. so that 360° may be suppressed as often as it can be, so far as the function of the angle is concerned. 83. The trigonometric functions of any angle whatever can now be readily expressed in those of an angle not exceeding 90°. Tims, 1. The trigonometric functions of any negative angle can be made to depend upon those of the corresponding positive angle (Art. 79). 2. Any angle exceeding 360°, as far as the trigonometric functions are concerned, may be replaced by an angle less than 360° (Art. 82). 3. Any angle exceeding 180° can in like manner be replaced by an angle less than 180° (Art. 80). 4. The trigonometric functions of any angle exceeding 90° may be made to depend upon those of an angle less than 90° (Art 77, 78). For example. BOOK II. 31 Bin 600° = sin (360° + 240°) = sin 240° = sin (180° + 60°) = —sin 60°, tan (—1000°) = —tan 1000°= —tan (720°+ 280°) — —tan 280° = —tan (180° + 100°) = —tan 100° = —tan (180° — 80°) = tan 80°. 84. It will be seen from the preceding articles that the sine and cosine may have any value between • — 1 and -\-l ; the tangent and cotangent, any value between — oo and -\-; the secant and cosecant, any value between — oo and — 1 and be- tween -\- 1 and -|- 00 . It might also be shown that the versed sine, coversed sine, and suversed sine may have any value be- tween and 2. It will also be found that no trigonometric function changes its sign except when it passes through the value zero or the value infinity. 85. The values of the functions of 0°, 90°, 180°, 270°, and 360° can be readily recalled, by being represented geometrically, according to the definitions of Art. 64. Thus, sin 0° = 0, cos 0° = 0^ = i2 = 1, tan 0° = 0, and sec 0° = OA — 1; sin90°= 0^' = 1, cos 90°= 0; sinl80°=0, cos180°=0j1' = — 1; sin 270°= 0^"'= —1, cos 270°= O. The tangent for either 90° or 270° would be a line drawn through A parallel to the secant, which would be A' A'" prolonged, and as they are each to be limited by their mutual intersection, they must both be infinite. The cotangent of either 0° or 180° would be an infinite line drawn through A', and parallel to the cosecant, which would be A A" infinitely prolonged. vers 0° = 0, vers 90° = vers 270° = ^0 = 1, vers 180° ^ A A" ^ 2 ; covers 0° = covers 180° := ^'0 = 1, covers 90° = 0, covers 270° = A' A'" = 2 ;• suvers 0° = A" A = 2, suvers 90° = suvers 270° =zA"0= 1, suvers 180° = 0. 32 TRIGONOMETRY. TABLE. Degrees. Sine. Cosine. Tangent. Cotangent. Secant. Cosecant. + 1 00 + 1 CO 90 + 1 00 00 +1 180 —1 00 — 1 CO 270 —1 00 CO —1 360 +1 00 +1 CO GENERAL FORMULA. 86. From the four fundamental formula (Art. 64), a large number of other formulae of general utility may be deduced. 87. To jind expressions for the products of sines and cosines, and for their sums and differences. The sum and difference of equations (17) and (19) are sin {a-\-b) -\- sin (a — 5) = 2 sin a cos h, (58) sin (ff-j- V) — sin (a — h) ^.2 cos a sin J ; (59) and the sum and difference of (18) and (20) are cos {a-\-h) -\- cos (a — i) = 2 cos a cos h, (60) cos {a-\-h) — cos (a — i) = — 2 sin a sin h. (61) Now, if in the formulae we let a-\-h^ A, and a — h^B, that is, a='2{A-\-B), and 6 = ^ (J. — ^), we shall have, sin vi -j- sin 5 = 2 sin ^ {A-\-B) cos ^ {A — B), (62) sin A — &m B = 2 coa ^ {A-\- B) sin ^ {A — B), (63) cos^ + cos5= 2cos ^ {A-\-B) cos ^ {A — B), (64) cos .B — cos ^ = 2 sin ^ {A-\-B) sin x {A — B), (65) in which A and B represent any two angles, and consequently admit of every possible value. These formulae are of frequent BOOK n. 83 application, especially in calculations effected by logarithms : (58), (59), (60), and (Gl) serve to transform a product to a sum or difference, and (62), (63), (64), and (65) serve to transform a sum or difference to a product. 88. Dividing formula (62) by (63), we have sin ^ -|- sin £ _ sin ^ (A + B) cos ^ (A — B) sin A — siuB ~^ cos"^ [a -\- B) sin | (A — B)' which, by means of (13) and (5), becomes sin A -\-sm B sin A — sin iJ ' or, : tan i (A^JB) cot J {A — B), and by (14), sin A -\-siaB tan ^ (A + B) sin A — sin fi tan ^ {A — B) ' (66) (67) sin A -\- sin B cot ^ (A — B) sin^^^sinTB cot I (A -f- B) ' that is. The sum of the sines of two angles is to their difference as the tangent of half the sum of the angles is to the tangent of half their difference, or as the cotangent of half their difference is to the co- tangent of half their sum. 89. By means of (62), (63), (64), and (13), we obtain, sin A + sin ^ ^2jin| (.4 +^)cosJ (A-B)^ cosA-\-co&B 2cos-^{A-^B)cos^(A — B) •'^ ' "\ / sin^-sin^ ^2cosj(^ + ^_2sin^ |U-^) cos^-)-cos-B icos^{A-\-B)cos^(A — B) ■^^ -" ^ '' that is, The sum of the sines of two angles divided h/ the sum of their cosines is equal to the tangent of half the sum of the angles ; and The difference of the sines of two angles divided hj the sum of their cosines is equal to the tangent of half the difference of the angles. 90. To find the tangent and cotangent of the sum of two angles h/ means of their tangents. Let A and B be the two angles ; then, by (13), tan (AA-B-\— '''"-(^^_+^) ■ tanC^-t-^)_^^^^^_^^^, 4 34 TRIGONOMETRY. or, substituting for sin {A -[-£), cos (A -j- B), their values from (17) and (18), , , , „N sin J. cos fi 4- cos yl sin B tan {A-\- B)=L =-^ — . — -j—. — 5. ^ ' ' cos A cos B — sm ^ sin U Dividing all the terms of the numerator and denominator by cos A- COS B, we have sin ^ i^sin B , , , ^. cos A '"cos-B tan (A-\- B) = -. — -. ^-5' ^ ' ' sin yl sm is cos A cos B or, by (13), / A \ T}\ tan ^ -j- tan B _„. tan (^ + ^) = ,_t,„^tanB ' ^^^^ and, by (5), ,,..„. 1 — tan A tan B ,_, . cot(^+^)= tan^+tani^ ' <^'^^ 91. To find the tangent and cotangent of the difference of two angles by means of their tangents. By (13), tan(^-^)="";^-g, ^ ■' cos (A — -B) then, in like manner as in Art. 90, we have, , , „, tan A — tan B tan {A — B)=: and "' 1+tan^tan.B' cot {A - ^ 1 -|- tan .4 tan B ' tan A — tan B (72) (73) 92. To find the sine, ^c. of double an angle, by means of the functions of the angle itself. In the expression for sin (^A-\-B) and cos (^A-\-B), let .5=^, then sin 2 ^ = sin ^ cos -4 -|- sin ^^ cos A, or, sin 2 ^ =: 2 sin A cos A ; (74) and cos 2 ^i = cos A cos A — sin .4 sin A, or, cos 2 ^ ::= cos* A — sin* A. (75) BOOK II. 35 Substituting in the latter, first the value of cos'' A and then the value of sin^^, from (9) and (11), we have cos 2 ^ = 1— 2 sin^^ (76) cos 2 4 = 2 cos^ A.— 1. (77) By means of (70) and (71), we have 93. To jind the sine, ^c. of half an angle hy means of the sine or cosine of the angle itself Let ^ A^^ A in (76) and (77), and transpose ; then 2 sin^ J ^ = 1 — cos ^, (80) 2 cos^ ^ A = 1 + cos ^ ; (81) whence . /I — cos J. . /I + cos A , , A — i/ ~ , cos ^ ^ = t / — L_ , (82) . sin i ^ /I — cos A , „, tan i ^ =: — ,—. =: 1 / ,— i T- (83) ^ COS ^ ^4 y 1 -|- cos ^ ^ ^ By multiplying both numerator and denominator by y'l — cos A, or by y'l -)- cos A, we obtain (12), . 1 — cos ^ sin ^ ... tan X A =z ; — = —- J. (84) ^ sm .4 1 -f- cos 4 ^- ^ NATURAL SINES AND COSINES. 94. Natural sines, cosines, &c. are the values of the sines, cosines, &c. expressed in natural numbers. 95. A table containing these values is called a table of nat- ural sines and cosines. 96. The semi-circumference of a circle whose radius is 1 is equal to 8.1415926 nearly (Geom., Prop. XV. Sch. 2, Bk. VI.), 36 TEIGONOMETEY. and this divided by 10800, the number of minutes in 180°, -will give .0002908882 for the arc of 1', which may be taken also for the sine of an angle of 1'. By means of formula (10), cos 1' = v^l — sin^ 1' = .9999999577. Then by transposition of formulas (58) and (60), sin (a-\-b) = 2 sin a cos b — sin (a — b), cos (a -j- S) = 2 cos a cos J — cos (a — b), in which, making b equal to 1', and a, in succession, equal 11 2', 3', &c., we obtain for the sines, sin 2' = 2 sin 1' cos 1' — sin 0' = .0005817764, sin 3' = 2 sin 2' cos 1' — sin 1' = .0008726646, sin 4' = 2 sin 3' cos 1' — sin 2' = .0011635526, &c., &c., and for the cosines, cos 2' = 2 cos 1' cos 1' — cos 0' = .9999998308, cos 3' = 2 cos 2' cos 1' — cos 1' = .9999996193, cos 4' = 2 cos 3' cos 1' — cos 2' = .9999993232, &c., &c., thus obtaining the sines and cosines up to 45°. The tangents may readily be found by dividing the sines by the cosines (13) ; and the secants, cotangents, and cosecants by dividing 1 by the cosines, tangents, and sines, respectively (Art. 51). 97. The sines, tangents, and secants of angles greater than 45° are respectively the cosines, cotangents, and cosecants of their complements, which are less than 45°; and, by their definitions, cosines, cotangents, and cosecants are the sines, tangents, and secants of complements (Art. 50). Thus, sin 46° = cos (90° — 46°) = cos 44°, tan 51° = cot 39°, cos 50° = sin 40, cot 88° = tan 2°. BOOK II. 37 Tables, therefore, do not go beyond 45°; or, rather, are so arranged that each number answers as a function of both ap angle less than 45° and its complement greater than 45°. TABLE OF LOGARITHMIC SINES, COSINES, &c. 98. A TABLE of LOGARITHMIC SINES, COSINES, &C. Contains the logarithms of the numbers expressing the natural sines, cosines, &c. 99. Since the sines and cosines are never greater than 1, and tangents likewise, when under 45°, their logarithms properly have negative characteristics. But to avoid the inconvenience of these, the characteristics are, by common consent, increased by 10. Thus the characteristic 9 is used in the place of — 1, 8 in place of — 2, &c. The radius, therefore, of the logarithmic sines, cosines, &c. is, as arbitrarily assumed, lO'", or 10,000,000,000. 100. In the accompanying table the degrees are given at the top and bottom of the page, and the minutes in the columns at the sides designated by M. The column headed D contains the increase or decrease for 1 second. This is obtained by taking one sixtieth of the difference between the logarithmic sine, cosine, &c. of an angle or arc, and that next exceeding it by 1 minute. The result is placed against the lesser angle or arc. To FIND THE Logarithmic Sine, &c. of ant Angle or Arc. 101. If the angle or arc is less than 45°, look for the degrees at the top of the table, and for the minutes on the left ; then, opposite to the minutes, on the same horizontal line, and in the column headed Sine, will be found the logarithmic sine ; in the column headed Cosine will be found the logarithmic cosine, &c. Thus, the logarithmic sine of 19° 23' is 9.520990, " " cosine of 31° 47' " 9.929442, " tangent of 43° 5' " 9.970922. 38 TRIGONOMETRY. 102. If the angle or arc is between 45° and 90°, look for the degrees at the bottom of the table, and for the minutes on the right ; then, opposite to the minutes, and in the column desig- nated at the bottom Sine, will be found the logarithmic sine ; in the column designated at the bottom Cosine will be found the logarithmic cosine, &c. Thus, the logarithmic sine of 80° 11' is 9.993594, " " cosine of 65° 69' " 9.609597, " " cotangent of 73° 35' " 9.469280. 103. If the angle or arc is between 90° and 180°, subtract it from 180°, and take the logarithmic sine, &c. of the remainder. Thus, the logarithmic sine of 112° is the logarithmic sine of 68°. " " tangent of 98° " " tangent of 82°. 104. If the angle or arc is expressed in degrees, minutes, and seconds, find the logarithmic sine, &c. of the degrees and min- utes as before ; then multiply the number opposite, in the column headed D, by the seconds, and add the product to the number first found, for sines and tangents, but subtract it for cosines and cotangents. Thus, if the logarithmic sine of 30° 25' 42" is required. The logarithmic sine of 30° 25' is 9.704395 Tabular difference, 3.59 Number of seconds, 42 Product, 150.78 150.78 Logarithmic sine of 30° 25' 42" is 9.704546 It is customary to omit the decimal figures at the right, bu*. to increase the last figure retained, by 1, when the figure at the left of those omitted is 5 or greater than 5. 105. The secants and cosecants are not included in the table, since they may be readily derived from the cosines and sines. By (5), sec A cos A = 1, and log sec A -\- log cos A = ; but as log sec and log cos are each increased by 10 (Art. 99), the second member of the equation must be increased by 20, that is, BOOK II. 39 logarithmic secant :^ 20 — logarithmic cosine, ^n like manner, logarithmic cosecant =20 — logarithmic sine. Hence, to find the logarithmic secant, subtract the logarithmic cosine from 20 ; and to find the logarithmic cosecant, subtract the logarithmic sine from 20. Thus, The logai-ithmic secant of 65° 59' is 10.390403 " " cosecant of 30° 25 ' 42 " " 10.295454. To FIND THE Angle or Arc corresponding to any Logarithmic Sine, &c. 106. Look in the column designated by the same name with the given logarithm for the sine, &c. which is nearest to the given one, and if the name be at the head of the column, take the degrees at the top of the table, and the minutes on the left ; but if the name be at ihefoot of the column, take the degrees at the bottom, and the minutes on the right. Thus, The angle or arc corresponding to the logarithmic sine 9.681443 is 28° 42'- The angle or arc corresponding to the logarithmic tan 9.984079 is 43° 57'- The angle or arc corresponding to the logarithmic cos 9.731603 is 57° 23 '. 107. If the given logarithmic sine, &c. is not found exactly, or very nearly, then, to find the seconds, subtract from the given logarithm that next less in the table, to the remainder annex two ciphers, divide the result by the number in the column headed D, and the quotient will be the number of seconds to be added to the degrees and minutes of the lesser logarithm for sines and tangents, or to be subtracted for cosines and cotangents. Thus, to find the angle or arc corresponding to the logarithmic sine 9.938070. Given log sine, 9.938070 Next less, 9.938040 corresponding angle, 60° 7' Diff. from column D, 1.21)30.00 25" The log sine 9.938070 has for its cor. angle or arc, 60° 7' 25' 40 TRIGONOMETKY. The angle or arc corresponding to the logarithmic tangent 9.497200 is 17° 26' 33". The angle or arc corresponding to the logarithmic cosine 9.792477 is 51° 40' 30". EXAMPLES. 1. Eequired the logarithmic sine of 28° 42'. Ans. 9.681443. 2. Required the logarithmic cosine of 59° 33' 47". Ans. 9.704657. 3. Required the logarithmic cotangent of 127° 2'. Ans. 9.877640. 4. Required the logarithmic sine of 81° 20'. Ans. 9.995013. 5. Required the logarithmic secant of 51° 40' 30". Ans. 10.207523. 6. Required the logarithmic tangent of 74° 21' 20". Ans. 10.552778. 7. Required the logarithmic cosecant of 102° 24' 41". Ans. 10.010270. 8. Required the logarithmic tangent of 1° 59' 61".8. Ans. 8.542587. 9. Required the angle of the logarithmic sine 9.999969. Ans. 89° 19'. 10. Required the arc of the logarithmic tangent 9.645270. Ans. 23° 50' 17". 11. Required the angle of the logarithmic cosine 9.598075. A-hs. 66° 39'. 12. Required the angle of the logarithmic cotangent 10.301470. Ans. 26° 32' 31". 13. Required the arc of the logarithmic sine 9.893410. Ans. 51° 28' 40". 14. Required the angle of the logarithmic cosine 9.421157. Ans. 105° 17' 29". 15. Required the arc of the logarithmic tangent 9.692125. Ans. 26° 12' 20"- 16. Required the angle of the logai'ithmic cotangent 9.421901. Ans. 75° 12' 6". BOOK III. SOLUTION OF PLANE TRIANGLES. 108. The solution of triangles is the process by which, when the values of a sufficient number of their elements are given, the values of the remaining elements are computed. The elements of every triangle are the three sides and the three angles. Three of these elements must be given, one of which must' be a side, in order to solve a plane triangle. The solution of plane triangles depends upon the following rUNDAMENTAL PROPOSITIONS. 109. In a right-angled triangle, the side opposite to an acute angle is equal to the product of the hypothenuse into the sine of the angle ; and the side adjacent to an acute angle is equal to the product of the hypothenuse into the cosine of the angle. Let ^ J5 C be a triangle having a right ■<^B 'P angle at G ; then, by (1), sin A, P ^4-1 sm 11 £= \ A therefore p = h sin A, l^=hsva.B. Again, by (4), COS A = -j- , cosi?=-f; therefore h = h cos A, p =z: A cos B. c (85) (86) 110. In a right-angled triangle, the side opposite to an acute angle is equal to the product of the other side into the tangent of the angle ; and the side adjacent to an acute angle is equal to the product of the other side into the cotangent of the angle. 42 TEIGONOMETKY. For, by (2), therefore Again, by (4), cot A therefore . p tan A^=i ~- 1 p=:b tan A, h_ h ^=^ p cot A, tan ^ = — , P b — p tan B. (87) oot B = f p — b cot B. (88) 111. In any plane triangle, the sides are proportional to the sines of the opposite angles. Let A B C he any triangle, in which the sides opposite the angles A, B, G, re- spectively, are denoted by a, b, and c. From one of the angles, as B, draw BD perpendicular to the opposite side A G, and denote the line B Dhj p. Then the A right-angled triangles, G BD, A B D, give, by (85), p =^ a sin G, jt> ^ c sin ^ ; whence, a sin C = c sin A, which gives the proportion a : c : : sin ^ : sin G. In like manner it may be proved that a : b : : s\n A : mi B, c : & : : sin C : sin ^, and these three proportions give a -.b : c : : sin A : sin B : sin G, which may also be written a h c sin B ' sin C (89) (90) (91) (92) (93) The angle G was acute, but had it been obtuse, or a right angle, the results would have been the same. The proposition, therefore, applies in every case. BOOK III. 43 112. In any plane triangle, the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference. For, by (90), a : b : : sin A : sin B ; whence (Geom., Prop. XII. Bk. II.), a -|- 5 : ct — 5 : : sin ^ -]- sin i? : sin A — sin B, which may also be written, a -^b sin ^ -|- ^™ -^ sin A — sin iJ ' a — b But, by formula (66), sin A -\- sin B tan ^ (.1 + B) _ therefore. sin A — sin iJ tan ^ (^1 — B) ' a 4- 5 _ tan ^ (yl + B) (94) a _ i — tan -^ (.1 — £) ' or, as it may be written, a -{- b : a — b :: ta,n .^ (A -}- B) : ta.n i (A — B). (95) 113. In any triangle, the square of any side is equal to the sum of the squares of the two other sides, diminished by ticice the rectangle of these sides multiplied by the cosine of the included angle. Let A B O he any plane triangle, in which the sides opposite the angles A, B, O, respectively, are denoted by a, b, c. Draw BD from one of the angles, B, perpendicular to the opposite side, AO. Then, if A is acute, we have by Geom- etry (Prop. XII. Bk. IV.), a' = V-^d' — 2b AD; but from the right-angled triangle A BD,hy (86), we have AD ^ c cos ^ ; therefore, a' — b^ -]- c'- — 2 be cos A, (96) 44 TRIGONOMETEY. When the angle A is obtuse, the point D will fall on the other side of A, and we have by Geom- etry (Prop. XIII. Bk. IV.), a^ — ¥ -{-€'-}- 2 b AD. But since BAD is now the supplement of D A 6 BAG, by Art. 77 we have AD = c cos BAD = — ccos BAG= — c cos A. Substituting this value oi AD, we have as before, a? =iiW -\- c^ — i he cos A. When ^ is a right angle and a the hypothenuse, cos^ is zero (30), and (96) becomes and thus the formula (96) is true, whatever the angle A m&y be. In like manner we have ¥=a^^c"' — 2 ac cos B, (97) c2 = a» -)- 52 _ 2 a J cos <7. (98) 114. The cosine of any angle of a plane triangle is equal to the fraction whose numerator is the sum of the squares of the contain- ing sides, diminished by the square of the opposite side, and whose denominator is twice the product of the containing sides. For, by (96), a^ = P -{- c" — 2 he cos A, whence, cos A = — ^^^ . (q^) 2 be ^ ' Similarly, from (97) and (98), we have cos ^ = "° +/'-*; cosC=^i^l (100) 2 ac ' 2 al) ^ ^ 115. By these formulEe the angles of a triangle can be found ivhen the sides are given, but they cannot be conveniently ap- plied in computation by logarithms. We then subtract both members of formula (99) from 1, and obtain 1 A 1 b' + c^-a" BOOK 111. 4") and, substituting for 1 — cos ^ its value, 2 sin^ J- A, by (80), we have 2 sin^ M = 1 -^-^/, ~^ = a'-Q>-cr ^ 2hc 2 be (a -\-b — c) (a — b -\- c) ~ 2hc ' whence, sin- i .4 =: - — ■ ^ ^ — ^. (101) Let now 2s = a-|-S-(-c, so that s is half the sum of thu sides of the triangle ; then a + J — c = 2 (s — c), a — b-\-c=2{s — b). Substituting these values in the preceding equation, and reduc- ing, we have sin J- ^ = t / — -b) (s — c) b c (102) n like manner we may obtain sin -^B= J^'-- -a) (.s — c) a c ' (103) sin ^ C = . /^™ -a) (s-b) n h (104) That is, The sine of half of any angle in a plane triangle is equal to the square root of half the sum of the three sides less one of the adjacent sides, into half the sum less the other adjacent side, divided by the rectangle of the two adjacent sides. 116. If 1 be added to both sides of (99), then, substituting for 1 -|- cos A its value, 2 cos^ ^ A, by (81), we have 2 cos'' i A= l-\ '—-, = -^^ — ^-4 . ■^ ' 2 be 2 be _ (b + c + a) (b + c — a) _ Whence, cos^ M = ^' ^ ' + ^V' + ' ~ - ^- a03> ■^ i be 46 TEIGONOMETKY. Let now s = half the sum of the sides of the triangle, as in Art. 115 ; then, b-\-c-\-a:=2s, h -\- c ■ — a = 2 (s — a). Substituting these values in the preceding equation, Ave have cosiA^i./{i'^L?:\ (106) Y he Similarly, cos A JB JiSL-^}, <107) V ac cos^C=./liEL£). (1(58) V ab That is. The cosine of half of any angle of a plane triangle is equal to the square root of half the sum of the three sides, into half the sum less the side opposite the angle, divided by the rectangle of the two adjacent sides. 117. Dividing (102) by (106), (103) by (107), and (104) by (108), we have, by (13), tan i ^ = . / .^l^'U^-^) ; (109) tan J ^ = . K' - «)_(^z:f ) ; (110) tan i C = 4 A'^H^HiE^ (111) V s {s — e) That is, Tlie tangent of half of any angle of a plane triangle is equal- to the square root of half the sum of the three sides, less 07ie of the ad- jacent sides, into half the sum less the other adjacent side, divided by half the sum, into half the sum less the side opposite the angle. SOLUTION OF EIGHT-ANGLED TRIANGLES. 118. In a right-angled triangle, the side opposite to the right angle is called the hypothenuse ; that adjacent to the right angle, and upon which the triangle is supposed to stand, is called the BOOK III. 47 hase ; and the other side adjacent to the right angle, the perpen- dicular. The base and perpendicular have been termed the sides about the right angle. Of the acute angles, that adjacent to the base has been termed the anglii at the base, and the other the angle at the perpendicular. Thus, let ^ 5 C be any right-angled triangle, with the right angle at G, then h represents the hypothe- nuse, h the base, p the perpendicular, A the acute angle at the base, and -S the acute angle at the perpendicular. U9. In order to solve the triangle, two elements other than the right angle must be given, one of them being a side. Hence there will be four cases in which there may be given, respectively, I. The hypothenuse and an acute angle. II. A side about the right angle and an acute angle. III. The hypothenuse and a side about the right angle. IV. The two sides about the right angle. Case I. 120. Given the hypothenuse and an acute angle. Let there be given, in the right-angled triangle ABC, the hypothenuse h and the acute angle A ; to find the angle B, the per- pendicular jo, and the base h. To Jind B. The angle B is the comple- A ment of A (Art. 44) ; hence, ^=90° — A To find p and h. By (85) and (8G) we have ^ =: ^ sin ^ ^ A cos B, h ^h cos ji = A sin B ; or, by logarithms, log jo =: log h -\- log sin A = log h -\- log cos B, (112) log h = \ogh -\- log cos A=^\ogh -\- log sin B. (113) 48 TEIGONOMETEY. That is, The logarithm of either side about the right angle is equa^ to the logarithm of the hypothenuse, plus the logarithmic sine of the opposite angle, or plus the logarithmic cosine of the adja- cent angle. Note 1. As the logarithmic sine and cosine are increased by 10 (Art. 99), the resulting logarithm will be so much too great, and must be diminished by 10. This increase by 10 will affect the work wherever the logarithms of trigo- \iometric functions are used. Note 2. The last figure of an answer may occasionally be found to differ from the one given in this work, when it has been obtained by the use o^ dif- ferent formulae or tables. The results are not, however, generally carried so far as to admit of such a difference. When two methods of solving give dif- lerent results, that is inserted which is most accurate, whether obtained by the nsual method or not. EXAMPLES. 1. Given the hypothenuse of a right-angled triangle equal to 1785.395 feet, and the angle at the base equal to 50° 37' 42"; \o solve the triangle. Solution. The angle at the perpendicular = 90° — 59° 87' 42'' — 30° 22' 18". Let, now, h =. 1785.395 feet and A — 50° 37'42", and we have, by (112) and (113), h = 1785.395 log 3.251734 log 3.251734 A — 59° 37' 42" log sin 9.935892 log cos 9.703813 ;?=: 1540.37 log 3.187626 5 = 902.708 log 2.955547 Ans. Angle at the perpendicular, 30° 22' 18" ; perpendicular, 1540.37 feet ; base, 902.708 feet. 2. Given the hypothenuse of a right-angled triangle equal to 25 yards, and one of the acute angles equal to 54° 30' ; to solve the triangle. 3. Given the hypothenuse of a right-angled triangle equal to 173.2 feet, and one of the acute angles equal to 37° 2' 43"; re- quired the other parts. Ans. Angle, 52° 57' 17" ; sides, 104.34 feet and 138.24 feet. BOOK III. 49 Case II. 121. Given a side about the right angle, and an acute angle. Let there be given (Fig. Art. 120) the side h and the angle A ; to solve the triangle. To find B. The angle B is the complement of A ; hence, 5 = 90° — A. To find p. By (87) and (88), we have p x^h tan j4 = 5 cot B ; or, by logarithms, logp = log h -(- log tan A = log h -j- log cot B. (114) To find h. By means of (113), we obtain log h = log h — log cos A = log h — log sin B. (115) Given the side p and the angle A ; to solve the triangle. To find B. We have, as before, the angle B, the complement of A or 5=90° — ^. To find h. By (87) and (88), we have b = p cot A^=^p tan B ; or, by logarithms, log b = log^ -|- log cot A = log jo -(- log tan B. (116) To find h. By means of (112), we obtain log h = log JO — log sin A = log^ — log cos B. (117) That is. The logarithm of either side about the right angle is equal to the logarithm of the other, plus the logarithmic tangent of the angle opposite, or, plus the logarithmic cotangent of the angle adjacent to the former. The logarithm of the hypotheniise is equal to the logarithm of either side about the right angle, minus the logarithmic sine of the angle opposite, or minus the logarithmic cosine of the angle adja- cent to the side. 5* 50 TRIGONOMETRY. EXAMPLES. 1. Given the side J of a right-angled triangle equal to 902.708 feet, and the acute angle A equal to 59° 37' 42"; to solve the triangle. Solution. The angle ^ = 90° — 69° 37' 42" = 30° 22' 18". By (114) and (115), we have J =902.708 log 2.955547 log 2.955547 ^ = 59°37'42" log tan 10.232078 ar. co. log cos 0.296187 jB = 1540.37 log 3.187625 ^=1785.395 log 3.251734 Ans. Angle B, 30° 22' 18"; perpendicular, 1540.37 feet ; hy- pothenuse, 1785.395 feet 2. Given one of the sides about the right angle of a right- angled triangle equal to 14<52 rods, and the opposite angle equal to 35° 30'; to solve the triangle. 3. Given the perpendicular of a right-angled triangle equal to 3555.4 yards, and the angle at the perpendicular equal to 33° 30' 47"; to solve the triangle. Ans. Angle at the base, 56° 29' 13"; base, 2354.4 yards; hypothenuse, 4264.3 yards. Case III. 122. Given the hypothenuse and a side about the right angle. Let there be given (Fig. Art. 120) the hypothenuse h and the side p ; to solve the triangle. To find A and B. By (1) and (4), we have sin A = cos i? = y- ; or, by logarithms, log sin A = log cos B = log p — log h. (118) 'To find b. By (85) and (86), we have b ^=h cos A=:h &m. B ; or, by logarithms, log & = log A -|- log cos ^ = log A -)- log sin B. (119) BOOK III. 51 Also, by Geometry (Prop. XI. Bk. IV.), we have h^ = p'^-{-V; (120) whence, V = h^ — jo'^ = (Ji -\- p) (h — p), h =^J[h-irp) (h-p); or, by logarithms, log * = i log {h+p)^i log (h — p). (121) That is, T/ie logarithmic sine of one of the acute angles, or the log- arithmic cosine of the other, is equal to the logarithm of the side opposite the former angle, minus the logarithm of the hy- pothenuse. The logarithm of either side about the right angle is equal to the logarithm of the hypothenuse, plus the logarithmic cosine of the angle adjacent, or plus the logarithmic sine of the angle opposite. EXAMPLES. 1. Given the hypothenuse -of a right-angled triangle equal to 1785.395 feet, and the perpendicular equal to 1540.37 ; to find the other parts. Solution. By (118) and (119), we have jB= 1540.37 log 3.187626 h = 1785.395 ar. co. log 6.748 266 log 3.251734 A = 59° 37' 42" log sin ) 9_935892 j cos A 9.703813 5=30" 22' 18" log cos i ° • h^ 902.708 log 2.955547 Ans. Base, 902.708 feet; angle at the base, 59- 37' 42"; angle at the perpendicular, 30° 22' 18". 2. Given the hypothenuse of a right-angled triangle equal to 73 feet, and one of the sides equal to 55 feet ; to solve the triangle. 3. Given the hypothenuse of a right-angled triangle equal to 643.7 rods, and the base equal to 473.8 ; to find the perpendic- ular and the two acute angles. A.V2,. Perpendicular, 435.73 rods ; acute angles, 42° 36' 12" 47° 23' 48"- 52 TEIGONOMETEY. Case IV. 123. Given the two sides about the right angle. Let there be given (Fig. Art. 120) the sides jo and h; to solve the triangle. To find A and B. By (2) and (4), we have tan ^ == cot i? = ^ ; or by logarithms, log tan A = log cot i? = log /) — log h. i;122) To find h. By (1), we have sin ^ = V-, whence h = — -— r ; (123) h ' sm ^1 ^ ' or, by logarithms, log h :^ log jo ■ — • log sin A ; (124) Also, by (120), A2 = / + W, whence h = ^f ^ b\ (125) That is. The logarithmic tangent of one of the acute angles, or the loga- rithmic cotangent of the other, is equal to the logarithm of the side opposite the former angle, minus the logarithm of the side adjacent. The logarithm of the hypothemise is equal to the logarithin of either side, minus the logarithmic sine of the angle opposite the side. EXAMPLES. 1. Given of a right-angled triangle the sidcp equal to 1540.37 feet, and the side b equal to 902.708 feet ; to solve the triangle. Solution. By (122) and (124), we have jo= 1540.37 log 3.187626 log 3.187626 h= 902.708 ar. co. log 7.044453 A = 59° 37' 42" loir tan ] ,r7„„_: '*'• <^°- '°g s'" ^ 0.064108 , 10.232079 Z?=30°22'18" log cot) ,_ 1785.305 j^g 3.251734 Ans. Hypothenuse, 1785.395 feet; acute angles, 59° 37' 42', 30° 22' 18". BOOK III. 53 2. Given the perpendicular of a right-angled triangle equal to 65 feet, and the ba.^e equal to 72 feet; to find the hypothenuse and the two acute angles. 3. Given the pei-pendicular of a right-angled triangle equal to 2.269 rods, and the base equal to 126.9 rods ; required the hy- pothenuse and the two acute angles. Ans. Hypothenuse, 126.92 rods; acute angles, 1° 1' 28", 88° 58' 32". SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 124. Since there must be given three elements, one of which is a side (Art. 108), there will be four cases, the data in them being, respectively, I. One side and any two angles. II. Two sides and an angle opposite one of them. III. Two sides and the included angle. IV. The three sides. Case I. 125. Given one side and two angles. Let there be given in the triangle ABO, the side «, and the two angles A and B; to solve the triangle. To find C. Since the sum of the three angles must be 180°, we have C= 180° — (^ + B). To find h and c. By (90) and (89), we have a : h : : An. A : sm B, a : c : : sin J. : sin C; whence, a sin B sin A a sin C sm A ' or, by logarithms, log S = log a -j- log sin B — log sin A, log c = log a -j- log sin C — log sin A. (126) (127) (128) 54 TRIGONOMETRY. That is, The logarithm of the required side is equal to the logarithm of the given side, plus the logarithmic sine of the angle opposite the re- quired side, minus the logarithmic sine of the angle opposite the given side. EXAMPLES. 1. Given of a triangle the side a equal to 9459.31 feet, the angle A equal to 71° 3' 34", and the angle B equal to 53° 26'; to find the sides 5, c, and the angle C. Solution. 0= 180° — (71° 3' 34" + 53° 26') = 55° 30' 26". Then, by (127) and (128), we have a= 9459.31 log 3.975859 log 3.975859 ^ = 71°3'34" ar.co.logsin 0.024176 ar.co.logsin 0.024176 5=53° 26' log sin 9.904804 <7= 55° 30' 26" log sin 9.91 6032 5 = 8032.28 log 3.904839 c = 8242.64 log 3.91 60OT Ans. Angle G, 55° 30' 26" ; side h, 8032.28 feet ; side c, 8242.64 feet. 2. Given one side of a triangle equal to 1 10 rods, the opposite angle equal to 50° 5', and an adjacent angle equal to 33° 55' ; to solve the triangle. 3.- Given one side of a triangle equal to 654 feet, one of the adjacent angles equal to 41° 0' 39", and the other adjacent angle equal to 55° 34' 8" ; to find the other parts. Ans. Angle, 83° 25' 13" ; sides, 432 feet, 543 feet. Case II. 126. Given two sides and an angle opposite one of them. Let there be given in any triangle, C A.B G, the two sides a, h, and the angle A opposite to one of them ; to solve the *^ triangle. To find B. By (90), we have A c 6 : : sin ^ : sin B, whence. sin B = h sin A (129) BOOK lU. 55 or, by logarithms, log sin J5= log b -\- log sin A — log a. (ISO) That is. The logarithmic sine of a required angle whose opposite side is git'en, is equal to the logarithm of that side, plus the loga- rithmic sine of the given angle, minus the logarithm of its apposite side. To find a We have 0= 180° — {A-\-B). To find c. By (128), after Cis found, we have log c =: log a -|- log sin O — log sin A. 127. Whenever the given angle is acute, and the side opposite to it is less tnan the side adjacent to it, there may be formed, as shown in Geometry (Prob. XI. Bk. V.), two triangles, each sat- isfying the given conditions, and, therefore, there will be two so- lutions. Thus ^^Fig. Art. 126) with two given sides a, h, equal respectively to t/ B and A C, and the given acute angle A oppo- site the less side C B, there may always be formed two triangles, ABC, A B' O, which have A, a, h in common, and the angles A B G, A B' G supplements of each other. In one of them, there- fore, the required angle is acute, and in the other it is obtuse. When the given angle is obtuse, the required angle must of necessity be acute, since a triangle can have but one obtuse angle. When the given angle is acute, and its opposite side is greater than the side opposite to the required angle, that must also be acute, since the greater angle must be opposite the greater side. When the side opposite the given angle is exactly a perpendic- ular let fall from C on ^ ^, the required angle is a right angle. If the side opposite the given angle be less than the perpendic- ular, the solution is impossible, since there will be no triangle with the given parts. When tv;o values are admissible for B, in case of ambiguity, two corresponding values will exist for G and c. 56 TEIGOKOMETRY. EXAMPLES. 1. Given of any triangle ABO, the side h equal to 216 yards, the side a equal to 117 yards, and the angle opposite the side a equal to 22° 37' ; to solve the triangle. Solution. By (130), we have a = 117 ar., co. log sin 7.931814 ft = 216 log 2.334454 ^ = 22° 37' log sin 9.584968 B = 45° 13' 55" or 134° 46' 5" log sin 9.851236 ^ -|-5= 67° 50 55" or 157° 20' 5" C = 180° — 67° 50' 55" = 112° 9' 5", or (7=180° — 157° 23' 6" = 22° 36' 55". Then, by (128), we have a = 117 log 2.068186 log 2.068186 C=112° 9' 5" log sin 9.966700 or=22°36'55"logsin9.584943 4 = 22°37'ar.co.logsin 0.415032 ar. eo. log sin 0.415032 f=281.785 log 2.449918 or =116.99 log 2.068161 Ans. Angle B, 45° 13' 55", or 134° 46' 5"; angle G, 112° 9' 5", or 22° 36' 56" ; side c, 281.785 yd., or 116.99 yd. 2. Given two sides of a triangle equal to 9459.31 feet and 8032.28 feet, and the angle opposite the first side equal to 71° 3' 34" ; to find the other parts. Solution. a = 9459.31 ar. co. log 6.024141 log 3.975859 5 = 8032.28 log 3.904839 ^ = 71° 3' 34" log sin 9.975824 ar. co. log sin 0.024176 £=63° 26' log sin 9.904804 0= 180° — 124° 29' 34" = 55° 30' 26" log sinj).916032 c = 8242.64 log 3.916067 Ans. Side, 8242.64 feet ; angles, 53° 26', 65° 30' 26". 3. Given two sides of a triangle equal to 80 rods and 142,6 BOOK III. 57 rods, and the angle opposite the second side equal to 96^ to solve the triangle. 4. Given in a triangle ABC, the side a equal to 32.1098 rods, tlie side b equal to 125.701 rods, and the angle A equal to 14° 48' ; to solve the triangle. Ans. Angle B, 90" ; angle O, 75° 12'; side c, 121.531 rods. 5. Given two sides of a triangle equal to 1540.37 feet and 760.9 feet, and the angle opposite the second equal to 30° 22' 8"; to find the other side and angles. Ans. Impossible. Case III. 128. Given two sides and the included angle. Let there be given in the triangle ABO B the sides a and b and the included angle 0, to solve the triangle. To jind A and B, we ha\e A -\- B =i 180° — G, and - A 6 C j- (^ 4- 5) = 90° — A C = complement of ^ C; whence, tan J (^ + ^) = cot -^ C. (131) Then, the half difierence oi A and B is found by means of (95), which gives ra -1- 5 : a — b:: i&n ^ {A -\- B) % Xaa } {A — B) ; whence, tan i (^ - 5) = ^J tan ^^ + i?) = ^ cot x G, (132) or, by logarithms, log- tan i {A—B) = log {a—b) — log {a-\-b) -f- log tan J. {A-\-B) = log {a — b)— log {a-\-b)-\- log cot | G. (133) That is, TOe hgarithmic tangent of half the difference of the two re- hired angles is egual to the logarithm of the difference of the 6 t)8 TRIGONOMETRY. given sides, minus the logarithm of their sum, plus the logarithmic tangent of half the sum of the required angles, or plus the loga- rithmic cotangent of half the given angle. Since ^ (A -\- ^) is known, when ^ {A — Ji) i> found, we A = i{A-^B) + i{A-B), B=iiA + B)-i(A-B). That is, The GREATER of the two required angles is equal to half their sum, plus half their difference ; and the smaller angle is equal to half their sum, minus half their difference. To find c. By (128), we have log c = log a -|- log sin G — log sin A. EXAMPLES. ' 1. Given of any triangle ABO, the side a equal to 9459.31 feet, the side h equal to 8032.28 feet, and the included angle equal to 55° 30' 26"; to find the side c and the angles A and B. Solution. A -\- B =^ 180° — O = 124° 29' 34", and J (^ + ^) = 62° 14' 47". Then, by (133), a + 6 = 17491.59 ar. co. log 5.757170 a — b= 1427.03 log 3.154433 }{A-\- B) = 62° 14' 47" log tan 10.278844 i; {A — B) = 8° 48' 47" log tan 9.190447 B = 53° 26' A = 71° 3' 34" ar. CO. log sin 0.024176 G = 55° 30' 26" log sin 9.916032 a = 9459.31 log 3.975859 c = 8242.64 log 3.916067 Ans. Side c, 8242.64 ft.; angle A, 71°3'34"; angle B, 53° 26'. 2. Given two sides of a triangle equal to 142.6 feet and 110 feet, and the included angle equal to 33° 55' ; to solve the tri- angle. BOOK III. 69 3. Given the two sides of a triangle equal to 153 rods and 137 rods, and the included angle equal to 40° 33' 12"; to find the other parts. Ans. Side,- 101.615 feet; angles, 78° 13' 1" and 61° 13' 47". Case IV. 129. Criven the three sides. Let there be given (Fig. Art. 128) the three sides a, h, and c ; to solve the triangle. To fina A, £, and G. By (102), (103), and (104), we have smM = ^- il y ac or, by logarithms, log sin i ^ = log(.— i)+ log (s-c)- log i- lege ^^3^^ log sin i 5 = log(..-a) + log(.-c)-loga-logc ^ ^^3^^ log sin i C= lo g (^ - ") + log Ji:^ ) - log a - log b _ ^^^^^ That is, Tlw logarithmic sine of half of any angle of a triangle is equal to the logarithm of the difference between half the sum of the sides and one of the adjacent sides, plus the logarithm of the difference between half the sum and the other adjacent side, minus the logarithms of those two sides, divided by 2. 130. A, B, and can also be determined by formulas (106), (107), and (108) for the cosine of half an angle, and by formute (109), (110), and (111) for the tangent of half an angle. "When the half angle is less than 45°, the table will determine it from its sine with greater precision than from the cosine, and tice versa when the half angle is greater than 45°. 60 TEIGONOMETRY The method by the tangent of half the angle is precise, and requires the use of but four logarithms. Note. This case may also be solved . by drawing a perpendicular from the A-ertex to the base of the triangle, thus dividing it into two'right-angled tri- angles, of which the hypothenuses are known, and the sum of whose bases is the base of the original triangle. Let s and s' represent CD and DA (Fig. Art. 113), then (Geom., Prop. XI. Bk. IV.), jo2 = c2 — s'2 = a^ — s^ or, s' — s'2 = a2 — c', whence, (s + s') (s — s') = (« + c) (a — c). Substituting b for s + s', s — s' ^ - " ^ ' , a form to which logarithms can be readily applied. Knowing s + s' and 5 — s', s and 5' can at once be found, and thence the angles A, C, and S, by Art. 122. EXAIIPLES. 1. Given of any triangle ABC, the side a equal to 216 yards, the side b equal to 217 yards, and the side c equal to 236 yards ; to find the angles A, B, and G. Solution. By (134), (135), and (136) we have a=216 ar.co.log 7.665546 ar.co.log7.665546 6 = 217 ar.co.log7.663540 ar.co.log7.663540 c = 235 ar.co.log 7.628932 ar.co.log 7.628932 «— a=118 log 2.071882 s— 6=117 log 2.068186 s—c= 99 log 1.995635 logJ..996635 2) 19.361995 log 2.071882 log 2.068186 2)19.356293 log sines 9.678147 2)19.469154 9.734577 9.680998 1 ^ = 28° 27' 47" ; i 5= 28° 40' 4".4 ; ^ C = 32° 52' 8".6. Ans. ^ = 5G°55'34"; .5= 57° 20' 8".8 ; C= 65° 44' 17".2. 2. Given the three sides of a triangle equal to 432, 543, and 654 ; to solve the triangle by means of the cosine. 3. Given the three sides of a triangle equal to 95.12, 162.34, and 98 ; to solve the triangle by means of the tangent. Ans. The angles, 32° 14' 53" ; 114° 24' 9" ; 33° 20' 58". BOOK IV. PRACTICAL APPLICATIONS. DETERMINATION OF HEIGHTS AND DISTANCES. 131. A Horizontal Plane is one which is parallel to the horizon. A Vertical Plane is one which is perpendicular to a horizontal plane. A Hokizontal Line is one which is parallel to the hori- zon. A Vertical Line is one which is perpendicular to a hori- zontal plane. 132. A Horizontal Angle is one the plane of whose sides is horizontal. A Vertical Angle is one the plane of whose sides is vertical. An Angle of Elevation is a verti- jj B cal angle having one side horizontal and the inchned side above it ; as the angle GAB. An Angle op Depression is a verti- cal angle having one side horizontal and the inclined side under it ; as the angle a" DBA. 133. To determine the height of a vertical object standing on d horizontal plane. Let B be the top of the object, and let it be required to find its height B G. 6* X i G2 TEIGONOMETSY. Measure from the foot of the object, in the horizontal plane, any convenient dis- tance, as ^ C, as a base line, and at A observe the angle of elevation GAB. Then, in the right-angled triangle ABC, we have known the side A C and the acute angle A ; therefore we can deter- mine the height B G hj Art. 121. EXAMPLES. 1. Standing on the edge of a moat 40 feet wide, I observe that the wall of a fort upon the opposite brink subtends an angle at the point of observation of 36° 52' 12"; required the height of the wall. Ans. 30 feet. 2. The angle of elevation of the top of a flag-staff, measured on a horizontal plane, at a distance of 89 feet from the foot of the staff, is 41° 29' ; what is the height of the staff? being D B ^ ^ ^ 134. To Jind the distance of a vertical ohject, its height given. Let B G he the object whose height is given, and let it be required to find the distance A G. Measure the angle of elevation GAB, or the angle of depression DBA, which is equal to GAB. Then, in the right- angled triangle A B G, we have known the side B G and the angles ; therefore we can find the distance A G hj Art. 121. EXAMPLES, 1. A tree 91 feet in height stands on the same horizontal plane with a dial, at which the angle of elevation subtended by the tree is 32° 22'; required the distance of the dial from the foot of the tree. Ans. 143.6 feet. 2. From the top of a house whose height is 30 feet, I observe that the angle of depression of an object standing on the same horizontal plane with the house is 36° 52' 12"; required the BOOK IV. 63 distance of the object from the base of the house, and also the length of the line that will just connect the object with the top of the house. 135. To Jind the distance of an inaccessible point on a hori- zontal plane. Let C be the point inaccessible from A and B, and let it be re- quired to find its distance from each of those points. Measure as a horizontal base line the distance between A and -Z?, and observe the horizontal a:^- glea C A B a.r>d G B A. Then, in ^ ^ the triangle ABC, there will be known the side A B and the angles ; therefore the sides A C and B can be found by Art. 125. EXAMPLKS. 1. Wanting to know the distances of two objects from a tree, inaccessible by reason of an intervening river, I measured the distance in a straight line between the two objects, and found it to be 772.45 feet ; I also found the horizontal angles formed by the extremities of the straight line with the tree to be 80° 58' 4" and 43° 33' 44". Required the distances of the objects from the tree. Ans. The one, 926.01 feet ; the other, 646.16 feet. 2. Two ships are engaged in cannonading a fort by the sea- side ; the ships are 131.89 rods apart, and the two angles at the ends of the straight line connecting the sliips, formed by that line and lines drawn to the fort, are 18° 52' 13" and 152°' 11' 42". Required the distance of each ship from the fort. 136. To Jind the height of an inaccessible object above a hori- zontal plane. First Method. Let B be the top of the object, and let it be I'equired to find the height B 0. Measure a horizontal base line, A C", of any convenient length, directly toward the object, and observe the angles of elevation at A and (7. Then, in the triangle A B C, since u TRIGOiNOMETEY. BOA is the supplement of C C' B, we have known the side A C" and all the angles ; therefore we can find the side AB by Art. 125. Then, in the right-angled triangle ABC, we have known the hypothenuse A B and the angles ; there- fore we can find the height BO by Art. 120. EXAMPLES. 1. Required the altitude of a hill whose angle of elevation, taken at the foot of it, was 55° 54', and 300 feet back, on the same horizontal plane with the foot, the angle was 33° 20'. Ans. 355.71 feet. 2. Two observers at sea, 800 yards apart, noticed at the same instant a meteor bearing due east from each ; to the one its angle of elevation was 57°, and to the other the same angle was 31° 28'. Required the altitude of the meteor above the horizontal plane of the ships. Second Method. Let B be the top of the object, and let it be re- quired to find the height B 0. Now, suppose it is not convenient to meas- ure a horizontal base line directly toward the object, and we measure it in any direction, A B', also meas- uring the angles GA B' and OB A. Then, in the horizontal triangle A B O, we know the side A B' and all the angles ; there- fore the side A Ccan be found by Art. 125. Then, also, by ob- serving the angle of elevation OA B, we shall, in the right-angled triangle ABO, know the side A O and all the angles ; therefore the height B O can be found by Art. 121. EXAMPLE. 1. A person on one side of a river observed an eagle's nest on an inaccessible mountain-crag on the opposite side, and beino- de- sirous of ascertaining its height above the level of the river, he measured along the shore a straight line 110 yards in length, and BOOK IV. 65 found the horizontal angles of its extremities with the object to be 38° 55' and 96°, and also the angle of elevation at the latter to be 45°. Required the height of the nest above the water. Ans. 240 feet. 137. To find the distance hetween two objects separated hy an impassable barrier. Let A and B be two objects separated by an impassable barrier, and let it be required to find the distance, A B, between them. Take any point, G, from which A and B are both visible and accessible. Measure CA and G B, and also note the angle A G B. ■ Then, since in the triangle A B G the two sides G A and G B, with their included angle', are known, the distance A B can be found by Art. 128. EXAMPLES. 1. Two bounds of a lot have between them an impassable mo- rass, and, wishing to find their distance apart, I have taken their distances from a third point, which could be seen from each. These distances are 124.75 and 171.41 rods, and the angle at that point subtended by the bounds is 99° 25'. How far are the bounds apart ? Ans. 227.91 rods. 2. The distance between two trees cannot be directly meas- ured, in consequence of an intervening obstacle, but within sight of each is a third tree, and their distances from this are known to be 274.65 and 396.11 yards, and the angle at that point sub- tended by the two trees is 8° 66' 5"- Eequired the distance between the two trees. 138. To find the distance between two inaccessible objects. Let C and D be the objects, and A and B two accessible points, from which both the objects are visible. Measure the base line A B, and observe the angles DAB, DBA, G A B, and G B A. Then, in the 66 TRIGONOMETRY. triangle DAB, since we have tlie side A B and all the angles, vre can find the side B D by Art. 125. In the triangle AB O Yie have the side A B and all the angles, hence we can find B C. Then, B D and B G being found, we have in the triangle BCD the sides BI) and B G, with their included angle ; therefore we can find the distance G Dhj Art. 128. EXAMPLE. 1. Wanting to ascertain the distance between a tree, Z*, and a flagstaff, C, on the opposite side of a river from me, I measured along the shore, on the horizontal plane with the objects, a base line, A B, of 110 yards. At A, the angle DAB equals 96°, and ; at B, the angle DBA equals 33° 55', 50'. Required the distance between the Ans. 261.81 yards. GAB equals 29° 56' and G B A equals 133° tree and the flagstaff. 139. To find the distances from a given point, of three oly'ecfs whose distances from each other are known. Let it be required to find the dis- tances from D, a given point, of three objects, A, B, and C, who-e distances from each other are known. Observe the angles A D G and BDG. Describe a circle about the tri- angle ADB, and draw A E and EB ; then the angle A B E is equal to the angle A D E, since both are measured by half of the same arc A E (Geom., Prop. XVIII. Bk. III.) ; also the an- gle B A E is equal to the angle B D E, for a like reason. Now, in the triangle A E B, the side A B and all the angles are known, hence the side A E may be found by Art. 125. Again, tlie sides of the triangle A B G being given, we may find the an- gle 5 ^ C by Art. 129 ; then, in the triangle AEG, tliere will be known the two sides A O, A E, and the included angle G A E, so that the angle AGE may be found by Art. 1 28. Then, in the triangle A G D, we shall know the side A O and the angles A G D and A D G ; therefore we can find the distance A D BOOK IV. 67 by Art. 125, and thence the other two distances, C D and B D. EXA3IPLE. 1. On approaching a harbor, at the point D, I observed three headlands, A, B, and C. Now it appeared from a chart that the distance from A io B was 800 yards, from AtoG 600 yards, and flora 5 to C 400 yards ; the angle A D C 1 found by obser- vation to be 33° 45', and the angle B D G tohe 22° 30'. What was the distance of each of the headlands from me? Ans. J, 710.19 yards; 5, 934.29 yards ; C, 1042.52 yards. DETERMINATION OE AREAS. 140. The Area of any figure, or its quantity of Furface, is de- ermined by the number of times the given surface contains some other area, assumed as the unit of measure, as a square inch, a square foot, &c. The areas oi parallelograms, triangles, trapezoids, &e. can be de- termined by direct application ,of the principles of Geometry ; but sometimes it is convenient to determine areas, especially of trian- gles, by means of their lines and angles, which requires the aid of Trigonometry. 141. To find the area of a triangle h/ means of two of and the included angle. Let A B Che any plane triangle, in which are given the sides h and c and the included angle A, to find the area of the triangle. Draw the perpendicular, p, from B 10 the opposite side, A G; then, since the area of the triangle is equal to half the product of its base by its altitude (Geom., Prop VI. Bk. IV.), area of A B G^ z ^ ?■ Put, by (85), p^c s'm A; its sides whence, area A B C:=.-^ b c &m A, (137) (138) 68 TRIGONOMETRY. or, by logarithms, log area A B C= log i 5 -|~ 'oS '^ + 'og ^™ ^- (139) That is, 77ie logarithm of the area of a triangle is equal to the logarithm of half of either side, plus the logarithm of either of the other sides, plus the logarithmic sine of their included angle. EXAMPLE. 1. Required the area of a triangle which has two of its sides equal to 105 and 85 feet, and the included angle equal to 28° 5'. Ans. 2100 sq. ft. 142. In like manner, the area of any parallelogram may be found, when two of its adjacent sides and the included angle are known ; for the diagonal divides a parallelogram into two equal triangles (Geom., Prop. XXXI. Cor. 1, Bk. I.). EXAMPLE. 1 . What is the area of a piece of ground, in the form of a parallelogram, which has two adjacent sides equal, respectively, to 120 and 212 rods, and their included angle equal to 85° 30'? 143. To Jind the area of a triangle hy means of a side and the angles. In the triangle ABO (Fig. Art. 141), let the side c and the angles be given, to find the area of the triangle. By means of Art. Ill we have 7 c sin B and by (85) p = c sin A. Substituting these values in (137), we obtain r- J 73 ^ c^ sin yl sin i? ^^ .-s areaof^i?C=-^-.-— , (141) or, by logarithms, log2 area^i?C=2 logc-j-logsin^-j-logsin^ — log sin C. (142) That is. The logarithm of double the area of a triangle is equal to twice the logarithm of either side, -plus the logarithmic sines of its adja- cent angles, minus the logarithmic sine of its opposite angle. BOOK IV. 69 EXAMPLKS. 1. A triangular lot has a side equal to 45 roils, and the adja- cent angles equal to 70° and 69° 40'; required (he area of the lot. Ans. 1378.41 sq. rods. 2. Given of a triangular field ABC, the angle A equal to 31° 27', the angle B equal to 101° 31', and the included side A B equal to 30 rods ; required the area of the field. 144. To find the area of a triangle hy means of its three sides. Let ABC (Fig. Art. 141) be the given triangle. Then, by (138), area of A B G = ^ be sin A; but, by taking twice the product of the values of sin ^ A and cos i- A, in. (102) and (106), we have, by (74), smA = ^\/s {s — a) (s — b) (s — c), (143) in wliich s denotes half the sum of the sides of the triangle. Substituting in the preceding equation this value of sin A, we obtain 2 - area oiABC^=.^bc X i" V'* (* — "■) (* — *) (* — '^) ' whence, area of ^ 5 C ^ V^s {s — a) {s — b) {s — c) ; (144) or, by logarithms, log area J5C = '°=-^ + '-°^(^=^^^+'°SjL-z*H^log(^-«). (145^ That is, The logarithm of the area of a triangle is equal to half the sum of the logarithm of half the sum of the sides and the logarithms of the remainders obtained by taking each side separately frmn half the sum of the sides. EXAMPLES. 1. Given the three sides of a triangle equal to 30, 40, and 60 rods, respectively ; required the area of the triangle. Ans. 533.27 sq. rods. 2. A certain fort is in the form of an equilateral triangle, whose sides are each 600 feet ; required the area occupied by the fort. 70 TRIGONOMETRY. MISCELLANEOUS PROBLEMS. « 1. The angle of elevation of a vertical tower is observed to be 30°, at the end of a horizontal base line of 100 yards, measured from its foot. Required the height of the tower. 2. A rope-dancer wishes to ascend a steeple 100 feet high, by means of a rope 196 feet long. If he can do so, find at what inclination he must be able to walk up the rope. 3. From the summit of a pier which rises 100 feet above the margin of a river, the angle of depression of the opposite margin was found to be 33° 16'. Required the width of the river. 4. If the distance of the moon from the earth be taken at 238500 miles, and the angle subtended by the semidiameter of the moon be iS' 33".5 at that distance, what is the moon's diam- eter? Ans. 2158 miles. 5. A point of land was observed by a ship at sea to bear east by south, that is, 11° 15' S. of E. ; and after sailing northeast 12 miles, it was found to bear southeast by east, that is, 33° 45' S. of E. Required the distance of the headland from the ship at the last observation. Ans. 26.07 miles. 6. From the top of Mont Blanc, 3 miles high, the angle of depression of the remotest visible point of the earth's surface is 2° 13' 27". Required the diameter of the earth, supposing it to be a perfect sphere ; and, also, the utmost distance from which the mountain is visible. Ans. Diameter, 7958 miles ; distance, 154.5 miles. 7. A side of the base of a square pyramid is 200 feet, and each edge is 150 feet ; required the slope of each face. Ans. 26° 34', nearly. 8. I have a meadow in the form of a parallelogram, whose two adjacent sides are 20 rods and 18 rods, including an angle of 78° 9' ; the same has been divided into two equal lots by a fence running diagonally. Required the area of each lot.: Ans. 176.16 square rods. 9; A traveler wishing to know the distance and heightof a mountain-top over which he had to pass, took the angle of its BOOK IV. 71 elevation at two stations, in a direct line towards it, the one 3 miles, or 5280 yards, nearer the mountain than the other, and found the angles to be 2° 45' and 3° 20'. Required the hori- zontal distance of the mountain-top from the nearer station, and its height. Ans. Distance, 24840 yards; height, 1447 yards. 10. From the top of a light-house the angle of depression of a ship at anchor was observed to be 4° 52', from the bottom of the light-houie the angle was 4° 2'. Required the horizontal distance of the vessel, and the height of the hill on which the light-house is placed, the height of the light-house being 60 feet. Ans. Horizontal distance, 4100.4 feet ; height, 289.12 feet. 11. When a tower 150 feet high throws a shadow 75 feet long upon the horizontal plane on which the tower stands, what is the sun's altitude (Art. 189) ? Ans. 63° 26' 6". 12. The sides of a triangle are equal to 3 and 12, respectively, and the included angle is 30° ; find, the hypothenuse of an equal right-angled isosceles triangle. Ans. 6. 13. From a window near the bottom of a house, which seemed to be on a level with the bottom of a steeple, I took the angle of elevation of the top of the steeple, equal to 40° ; then from another window, 18 feet directly above the former, the like angle was 37° 30'. Required the height and distance of the steeple. Ans. Height, 210.4 feet; distance, 250.8 feet. 14. Two pulleys, whose diameters are 6 inches and 4 feet 3 inches, respective^, are placed at a distance of 3 feet 6 inches from centre to centre. What must be the length of a belt which shall connect them, by passing around their circumferences, with- out crossing? Ans. 15 feet 5.9 inches. 15. A tower is surrounded by a circular moat. At noon on a certain day, the shadow of the top of the flag-staff is observed to project 45 feet beyond the edge of the moat. When the sun is due west, on the same day, the shadow projects 120 feet beyond the moat. The distance between the extremities of the shadows is 375 feet. The angle of elevation of the top of the flag-staff from any point of the edge of the moat is 60°. Find the height of the tower, and the altitude of the sun at noon. Ans. 311.77 feet; 54° 10' 57"- BOOK V. SPHERICAL TEIGONOMETEY. DEFINITIONS. 145. Spherical Trigonometry treats of methods of com- puting spherical triangles. 146. A SPHERICAL TRIANGLE is a portion of the surface of a sphere bounded by three arcs of a great circle, each of which is less than a semi-circumference. The three planes in which the arcs lie form a polyedral angle at the centre of the sphere. The ANGLES of a spherical triangle are the diedral angles made by the plane faces which form the polyedral angle. 147. The sides and angles of spherical triangles are usually both expressed in degrees, minutes, &c. The circumference, however, is sometimes supposed to be divided into 24 equal parts, called hours; each hour into 60 equal parts, called minutes of time ; each minute into 60 equal parts, called seconds of time. Then a side is expressed by the number of hours, minutes, seconds, and decimal parts of a second, which it contains. Hours, minutes, and seconds are denoted by h., m., and s. Thus, 3h. 35m. 6,8s. RELATIONS BETWEEN THE SIDES AND ANGLES OF SPHERICAL TRIANGLES. 148. In any spherical triangle, the sines of the sides are pro- ■jiortioncd to the sines of the opposite angles. BOOK V. 73 Let ^ jB C be any spherical triangle ; A, £, and C the angles opposite to its sides ii, b, and c, respectively ; and the centre of the sphere. Take any point B in B, and draw B D perpendicular to the plane A G ; from D draw D A', D C", perpendicular to A, C, respectively ; join B'A', B C". B' C is. a right angle (Geom., Prop. VI. Bk. VII.) ; therefore, B G'= OB sin B OG' = OB sin a, and B D= B G< sin B C D = B C" sin G=OB sin a sin G. In like manner, B D= OB' sin e sin A ; and, by the two preceding equations, B sin a soi. G ^ B sin c sin A, sin a sin A whence. sin C" sm c or, in the form of a proportion, sin o : sin c : : sin A : sin G. In a similar way it may be proved that sin a : sin 5 : : sin A : sin B, sin c : sin 5 : : sin C : sin B. (146) (147) (148) (149) The figure supposes a, c, B, G, &c. each less than 90°, but the relation stated may be shown to hold when the figure is modified to meet any case whatever. For instance, if G alone is greater than 90°, the point D will fall beyond G, instead of between OG and OA; then, B'GD will be the supplement of G, and thus, since the sine of an angle and its supplement are the same, the sine of B' CD is still equal to the sine of G. 7* 74 trigonomp:try. 149. In any spherical triangle, the cosine of any side is equal to the product of the cosines of the other two sides, plus the product of the sines of those two sides into the cosine of their included angle. Let ABC he any spherical tri- angle, the centre of the sphere. Draw the plane B' A' O per- pendicular to A. Then the an- gle B'A'C is equal to the angle A, the angle B' C" measures the side a, and in the triangles A' B' C", B O we have, by Art. 113, B' 6" = A> B'-^A; C'—'i A' B' X A' O cos A, B' V =0 B'-\- C" — 2 B' X C cos a. Subtracting the first equation from the second, observing that OH' — A'B' and OlJ' — A' C are each equal to (TJ', since the triangles A' B', O A' C are right-angled at A', we have 0=2 Oj'+ ^ A' BX A' C cos ^ — 2 B X O cos a; ,, . OA'XOA' A'B'XA'C . therefore, cos a = o-j7x~0-C" + B' X C ""^ ^- Substituting the functions derived from the triangles A' B, A' C, we have cos a = cos b cos c -j- sin b sin c cos A. {^^^) In like manner may be deduced cos b = cos c cos a -\- sin c sin a cos B, (1^1) cos c = cos a cos b -\- sin a sin b cos 0. (152) The preceding construction supposes the sides b and c, which contain the angle A, to be both less than 90°, but the formulas obtained may be shown to be applicable in all cases. BOOK V. r5 150. In any spherical triangle, the cosine of any angle is equal to the product of the sines of the other two angles into the cosine of their included side, minus the product of the cosines of those two angles. Let A'B'C be the polar triangle o^ A B C; denote its angles by A', £', and C", and its sides by a', h', and c'. Then (Geom., Prop. IX. v i Bk. IX.), ^'=180" — a, ^=180° — J, 0-- = 180"- :180°- -c; ■C. a' = 180°— ^ V=\%(f — B, Applying (150) to A'BC, we have cos a' =^ cos 5* cos c* -|- sin b' sin c" cos A' ; or, by (46), — cos ^ = cos 5 cos C — sin ^sin Ccos a; whence, cos ^ = sin B sin C cos a — cos BiMs O. (153) In like manner may be deduced cos ^=sin G sin A cos b — cos Ccos A, (1^4) cos <7^ sin A sin B cos c — cos A cos B. (155) 151. In any spherical triangle, the cotangent of one side into the sine of another side is equal to the cotangent of the angle op- posite the first side into the sine of the included angle, plus the cosine of the second side into the cosine of the included angle. By (150) and (152) we have cos a = cos 6 cos c -|- sin i sin c cos A, cos c = cos a cos 5 -|- sin a sin b cos C; and by means of (147), sin C sin A' Substituting these values of cos c and sin c in the first equation, we obtain 76 TRTGONOMKTRY. ^ ,. sin a sin ft cos A sin C COS a ^ (cos a COS 6 -j- Sin a sm cos U) cos fr-| ^m^4 ' or, cos o = cos ffl cos^ 5 + sin a sin J cos 5 cos C + sin a sin J cot A sin C. Therefore, transposing cosacos^J, and observing that, by (11), cos a — cos a cos^ J ^ cos a sin^ b, we have cos a sin^ i ^ sin a sin b cot ^ sin C + sin a sin b cos b cos 0, and dividing the whole by sin a sin b, we obtain cot a sin J = cot ^ sin C -f- cos J cos C. (156) 152. By interchanging the letters in (15G), we obtain cot a sin c = cot A sin JB -\- cos c cos B, (157) cot b sin a = cot 5 sin C -|- cos a cos O, (158) cot 5 sin c ^ cot B sin ^ -[~ <"""' ^ '^os ^, (159) cot c sin a = cot C sin B -{- cos a cos i?, (160) cot c sin 6 = cot C sin ^4 -j- cos b cos A. (161) 153. The formulte developed in the preceding articles are general, and apply to every case of spherical triangles, but re- quire some transformations to render them more convenient for logarithmic computations. The formulae (150), (151), and (152) of Art. 149 are consid- ered the fundamental formula of spherical trigonometry, since from them all its other formulae may be deduced. 154. To express the sine, cosine, and tangent of half an angle of a triangle as functions of the sides. By means of (150) we have . cos a — cos b cos c ,^ „„i cos ^ = • I • ) (162) sm sin c but this formula is not suited to logarithmic computation. We then subtract each member of the equation from 1, and obtain (Art. 63), . , cos a — cos b cos c cos (b — c) — cos a 1 cos ^ ^ 1 ■ , ■ ^ -^ ^^^ tA . sin sm c sm o sin c BOOK V. 77 Substituting for 1 — cos ^ its value, 2 sin^ ^ A (80), we obtain -, . , , , cos (h — c) — cos a 2 sin^ J- J. = ^^ — pA . sin sin c Now if, in (65), we make A =: a, and 5 = b — c, i(A-^B) = i{a + b-c), i{A~B)^i{a-b-\-c), then, cos (J — c) — ■ cos a ^ 2 sin J (a -j- J — c) sin ^ (a — b-\- c), which, substituted in the preceding equation, gives sin sin c ^ ■^ Let, now, s = half the sum of the sides of the triangle ; then, a -)- J — c = 2 (s — c), a — b-j-c = 2(s — b). Substituting these values in the last equation, and reducing, we have , . /sin (s — b) sin (s — c") sm i ^ = 1 / 5^ — ^^r—. — !^ i-. y sin sm c (164)' Similarly, mniB = J'^''^'-''^-']^^'-''\ y sin c sm a (165) sin^ ^^ Ain(s-«)sin(.s.-J)_ y sin a sin o (166) Adding each member of equation (162) to 1, and observing that 1 -j- cos ^ = 2 cos^ ^ A (81), by means of (65), we have eog2 xA = ^'" -H" + ^ + g) sin 1(6 4- c — g ) ^ sin J sin c Introducing i = ^ (a -[- * + '^)! and reducing, we have , , /sin s sin (s — a) cosiA = ^/ ^-jA ^. (167 y sin sm c ^ ■' „. ., , , r> /sin s sin (s — b) Similarly, cos^^ = ./ -. >; i, (168) '' y sm c sm a ^ ■' , ^ /sin s sin (s — c) ,-, ^„n cos A C —'. 4 / i '■■— ,-— • (169) y sin a sin S ^ ' 78 TRIGONOMETRY. Again, dividing (164), (165), and (166) by (167), (168), and <'169), respectively, we obtain ^^^ / sin (.-6) sin (.- g •' y sin s sin (s — ") tan i 5 = ./ S-9f (^-f °)" , (171) -^ y sin s sin (s — w) ■^ y si: tan 4-0=: sin (,s — a) sin (s — V) . , , . (172) sin « sin \s- — c) 155. To express the sine, cosine, and tangent of half a side of a trianyle as functions of the angles. By means of (153) we have cosvl -|-cosScosC sin B sin C (173) Whence, cos A -\- cos B cos C cos A -\- cos (B -\- C) 1 — ^cosra= 1 ' — ■ ■ ^nd sin^ J- a == ■ sin B sin C sin B sin C cos .^ (^ -f- £ +C) cos ^ (g + C — A) (174) sin £ sin C Making ^ {A -\- B -\- C) ^ S, substituting, and reducing, we have •,Scos(S — A) J« = y/- sin B sin C Also, sin i J = / — cos5c os (S-£) ^ V sin C sin 4 sin i c = /-cosSc os_(^--C) _ V sin A sin B In like manner, we obtain cos i « =:. / cos(5- ^)cos(-5^rc) ^ y sin £ sin C ,os i J = / -^(S^Cyco^(S-=^A) ^ V sin C sin A cos J c = /cor(5-^)cos'(>S-:B) _ y sin A sin B (175) (176) BOOK V. 79 Hence, o tan A a = V / -= ''°11_''^1^lA] \ COS (S — B) cos (S — C tan I b — / - cosT^^("5 -'B) V cos C^' — C) cos C-b'— ^) tan A c = / -cos^cos(5- Cr •^ Vco3(S — ^)cos(S — fi) (177) Since S is always greater than 90° and less than 270° (Geom., Prop. X. Bk. IX.), cos S is always negative, and therefore — cos S in the numerators of the first and third of the above sets of formulae is essentially positive. 156. To prove Napier's Analogies. sin A Let then, sin B sin b ' (178) (179) sin a sin ^ = m sin a, sin 5 = m sin h, sin A -\- sin B^m (sin a -\- sin i), sin A — sin £ =^ m (sin a — sin h), By (153) and (154) we have cos A -)- cos B cos C = sin 5 sin Ccos a=^m sin C cos a sin b, cos ^ -|- cos A cos C:=: sin ^ sin C cos b^m sin C sin a cos 5. Adding these equations, factoring, and reducing by (17), (cos A -\- cos B) (1 -|- cos G) :^ m sin C sin (a -|- b). Dividing (178) by (180), and multiplying by sin G, (180) sin A -\- sin B X sin C sin a -|- sin J cos A -\- cos £ 1 -|- cos C sin (a -|- 6) Now, by means of (62), (63), and (74), we obtain sin a -)- sin J cos ^ (a — 6) sin (a -\-b) cos ^ (a -{-b)' (181) (182) 80 TEIGONOMKTKY. , sin a — sin b sin A (a — b) ,- „„> and —. — 7 — |— =^ ^ -T — — — r~iA' {'■°'^) sin (a -(- *) sin | (a + *) Substituting in (181) the value of each expression, from (68), (84), and (182), cos ^ (a -\- b) cot ^ C cos"! (a^— J) tan ^(^+5) ' In like manner, from (179) and (180), sin A — sin 5 sin C sin a — sin b cos A -\- cos B 1 -\- cos C sin {a -\- b) whence, by (69), (84), and (183), , , 1 T.N , ^ sin i (a — b) tan H^ - -B) X tan i 0= -;~|-|^ , or, sin ^ (a -\-b) cot | C sin I (a — b) tan ^(A—B)' (184) (185) Formulae (184) and (185) may be thus expressed: ooa ^ (a + J) : cos J. (a — 5) : : cot ^. C : tan i {A-{- B), (180) sin ^ (a-j- J) : sin i (a— S) : : cot^ C: tan J- {A — B). (187) That is, The cosine of half the sum of two sides of a spherical triangle is to the cosine of half their difference as the cotangent of half the included angle is to the tangent of half the sum of the other two angles. The sine of half the sum of two sides of a spherical triangle is to the sine of half their difference as the cotangent of half the in- cluded angle is to the tangent of half the difference of the other two angles. By applying (186) and (187) to the polar triangle. Art. 150, tve obtain BOOK V. 81 COS i {A-^JB) i cos J (^ — 5) : : tan ^ c : tan ^ (a + b), (188) sin i (A-'f-B) : sin ^ (A — B) :: tan i c : tan ^ (a — b). (189) That is, 77ie cosine of half the sum of tiro angles of a spherical triangle is to the cosine of half their difference as the tangent of half the included side is to the tangent of half the sum of the other two sides. The sine of half the sum of two angles of a spherical triangle is to the sine of haf their difference as the tangent of half the in- cluded side is to the tangent of half the difference of the other two sides. The above four proportions are called, from their inventor, Napier's Analogies. RELATIONS BETWEEN THE SIDES AND ANGLES OF EIGHT-ANGLED SPHERICAL TRIANGLES. 157. 27ie sine of either oblique angle is equal to the sine of the opposite side, divided h/ the sine of the hypothenvse. Let A B C he any spherical trian- B gle, right-angled at C. By means of (146) we have , 0111 Jf . ^ sin A = -.^ sin C ; sin n sm jo sin h but, as C == 90°, sin C = 1 , and sin A In like manner. sin B -- sin b sin h' (191) 158. The cosine of either oblique angle is equal to thv tangent of the adjacent side, divided hy the tangent of the hypothenuse. By means of (161) we have cot h sin b = cot C sin A -j- cos b cos A ' but, if (7= 90°, then cot (7=0, and 8 82 TRIGONOMETRY. cot A sin J = cos h cos A, . cot A sin J ^ , ^ , or, cos A = ; — = cot A tan b : cos whence, cos A = j. (192) tan h ^ ' Also, by means of (160), cos B = ^^-l (193) tan h ^ ' 159. The tangent of either oblique angle is equal to the tangent of the opposite side, divided by the sine of the adjacent side. By means of (156) we have cot p sin b = cot ^ sin C -j- cos b cos G; but, by making G^ 90°, sin (7=1, cos 0^ 0, and sin b cot A = cot p sin b = tanp' or, tan^=:*4^. (194) sin ^ ^ Also, by means of (158), tan 5=^. (195) sin ^ ^ 160. The sine of either oblique angle is equal to the cosine of the other, divided by the cosine of its opposite side. By means of (154) we have cos -S :;= sin ^ sin G cos h — cos A cos G, which, by making G = 90°, becomes cos B = cos b sin A, , . . cos B ,. „ „x whence, sm A = ~. (196) cos ^ ' In like manner, by means of (153), •r. cos A sm B = . (197) cos p ^ ' 161. The cosine of the hypothenuse is equal to the product of the cosines of the other two sides. BOOK V. 83 By means of (152) we have cos h = cos p cos b -j- sin p sin b cos O, which, by making C =^ 90°, becomes cos h = cos p cos b. (198) 162. The cosine of the hypothenuse is equal to the product of the cotangents of the two oblique angles. By means of (155) we have cos G = sin A sin B cos h — cos A cos B, which, by making G = 90°, becomes sin A sin B cos h = cos A cos B, cos A cos B cos h : sin A sin B ' cot j4 cot -B. (199) 163. The preceding formulse may readily be remembered from their similarity to the corresponduig ones for plane triangles ; and, for convenience of reference, they are brought together in the fol- lowing TABLE. 1. 3. 5. 7. 9. , sm p sin A =^ -r-^. sm A cos A : tan b tan h , tan » tan A = - sin A = sm b' COS B cos b cos h = cos p cos 5. 2. 4. 6. 8. 10. sin B := cos B = tan 5 = sin 5 := sin b sin A' tan^ tan h tan 6 sin p' cos *4. cos p cos A = cot A cot 5. SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES. 164. The solution of spherical triangles is the process by which, when the values of a sufficient number of their six elements are given, we calculate the values of the remaining elements. 84 TRIGONOMETRY. In order to solve a right-angled spherical triangle, two of its elements, other than the right angle, must be given. 165. The formula requisite for the solution of right-angled spherical triangles are readily furnished by means of tlie rela- tions demonstrated in the foregoing articles. Thus, sin A ^= -. — ^ gives sin jo = sin ^ sin k, (200) sin B = ^!— I " sin b = sin B sin h, (201) sm ft ' \ / cos A = 7 — - " cos A ^ cot h tan b, (202) cos B = ^1^^ " cos B=coth tan p, (203) tan ^ = ^^- " sin » = cot j5 tan b, (204) sin ^ ' » V / . tan p . , . ^„„,, tan A = -. — ' " sm o = cot ^ tan p, (205) cos A sin B = " cos ^ 1= sin i? cos p, (206) cos p J ' \ / sin A = -. — T " cos i? = sin ^ cos 5, (207) cos ' \ / which, with equations (198) and (199), cos h = cos JO cos b, cos h = cot A cot B, enable us to determine every case of right-angled spherical tri- angles. For every one of these ten equations is a distinct com- bination, involving three of the five quantities, p, b, h, A, B\ and five quantities, taken three at a time, can be combined only in ten different ways. Napier's Circdlar Parts. 166. If, in any right-angled spherical triangle, the right angle be left out of consideration, the two sides adjacent to the right angle, and the complements of the hypothenuse and of the two ether angles, are called the five circular parts of the triangle. BOOK V. ^5 Thus, in the spherical triangle ABC, right-angled at C, the cir- cular parts are p, h, and the com- plements of h. A, and B. 167. When any one of the five parts is taken for the middle part, the two adjacent to it, one on either side, are called the adjacent parts, and the other two parts ai'e called the opposite parts. Then, whatever be the mid- dle part, we have as The Rules of Napier. I. The sine of the middle part is equal to the prodtKt of the tangents of the adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. 168. Napier's rules may be proved by showing that they agree with the results already established. Art. 165. Thus, 1. Let b be taken for the middle part ; then p and the com- plement of A will be the adjacent parts, and the complenients of B and h will be the opposite parts, and by the rules we have sin b = tan (com. A) t&n p, sin b = cos (com. £) cos (com. h) ; whence, by Art. 50, sin b = cot A tan jB, sin 5 :^ sin B sin A which agree with (205) and (201). In like manner, if p be taken as the middle part, sin p = tan (com. 5) tan b, sin p = COS (com. A) cos (com. h) ; whence, sin p = cot B tan b, sin p = sin A sin h, H'hich agree with (204) and (200). 2. Let the complement of h be taken as the middle part ; 86 TRIGONOMETRY. then the complements of A and £ will be the adjacent parts, p and b the opposite parts, and we have sin (com. h) = tan (com. A) tan (com. B), sin (com. h) = cos p cos b ; whence, cos A = cot ^ cot B, cos A ^^ cos p cos b, which agree with (199) and (198). 3. Let the complement of A be taken as the middle part ; then b and the complement of h will be the adjacent parts, p and the complement of -B the opposite parts, and we have sin (com. A) ^ tan (com. /*) tan b, sin (com. A) = cos (com. B) cos p ; whence, cos A = cot h tan b, cos ^ = sin jB cos p, which agree with (202) and (206). In like manner, sin (com. B) = tan (com. h) tan p, sin (com. B) = cos (com. A) cos b ; whence, cos B ^ cot h tan p, cos ^ :^ sin ^ cos 5, which agree with (203) and (207). 169. Any element of a spherical triangle is less than 180° (Geom., Art. 505, 539). Two parts are said to be of the same species when they are in the same quadrant, that is, when they are both less, or both greater, than 90° ; and of different species when one terminates in the first and the other in the second quadrant. 170. In order to determine whether a part sought is less or gi-eater than 90°, the algebraic signs of the terms should be ob- served, according to Art. 68 or 78. When, however, the part sought is determined by its sine, since the sines in both the first and second quadrants are positive, there will be two solutions, BOOK V. »( unless the ambiguity be removed by one of the following rules : — 1. In any right-angled spherical triangle, an oblique angle and its opposite side are always of the same species. For, by (205), sin 5 = cot A i&np, in which, since sin h is always positive, cot A and tan p must always have the same sign, that is, A and p must be of the same species. 2. When the two sides about the right angle are of the same species, the hypothenuse is less than 90°, but when they are of different species, the hypothenuse is greater than 90°. For, by (198), cos /* = cos p cos b, in which, if cos p and cos b have the same signs, cos h will be positive, but if they have unlike signs, cos h will be negative. 171. In the solution of right-angled spherical triangles, there will be six cases to consider, in which there may be given, re- spectively, I. The hypothenuse and an oblique angle. II. The hypothenuse and one side. III. One side and its adjacent oblique angle. IV. One side and its opposite oblique angle. V. The two sides about the right angle. VI. The two oblique angles. Case I. 172. Given the hypothenuse and an oblique angle. Let there be given in the right- angled spherical triangle A BC, the hypothenuse h and the oblique angle A ; to solve the triangle. To find p. Make p the middle part, and we have, by Napier's rules, or by (200), sin p = sin A sin h. 88 TRIGONOMETRY. or, by logarithms, log sin p = log sin A -\- log sin k. (208) To find h. Make the complement of A the middle part, aad we have, by Napier's rules, or by (202), cos A = cot h tan h ; whence, tan h = tan h cos A, (209) or, by logarithms, log tan h = log tan h -\- log cos A. (210) To find B. Make the complement of /* the middle part, and we have, by Napier's rules, or by (199), cos h = cot A cot B ; whence, cot 5^ cos A taxi A, (211) or, by logarithms, log cot B = log cos h -|- log tan A. (212) Thus, b and B, by observing the algebraic signs, are deter- mined without ambiguity ; and p, though determined by its sine, is not ambiguous, since it must be of the same species as A (Art. 170). EXAMPLES. 1. Given in a right-angled spherical triangle ABC, right- angled at O, the hypothenuse h equal to 105° 34', and the angle A equal to 80° 40' ; to solve the triangle. Solution. By (208), By (210), By (212), A,logsin+9.983770 log tan— 10.555053 log cos— 9.428717 /l,logsin-j-9 .994212 Iogcos-4- 9.209992 log tan-|-l 0.784220 7>,logsin+9.977982 S,logtan— 9.765045 i?,logcot— 10.212937 Hence, _p = 71° 54' 33", J= 149° 47' 37", B=: 148° 30' 54". 2. Given in the spherical triangle ABC, right-angled at (7, BOOK V. 89 the hypothenuse h equal to 70° 23' 42", and the angle A equal to 66° 20' 40" ; to find the other parts. AnB. p, 59° 38' 26"; b, 48° 24' 15"; B, 52° 32' 55". Case II. 173. Given the hypothemise and one side. Let there be given (Fig. Art. 172) the hypothenuse h and tha «ide p ; to solve the triangle. To find A. Make p the middle part, and we have, by Napiers rules, or by (200), sin p = sin A sin h ; whence, sin A = . , , f213") sin A ^ ' or, by logarithms, log sin A = log sin p — log sin h. (214) To find B. Make the complement of B the middle part, and we have, by Napier's rules, or by (203), cos B = cot h tan p, or, by logarithms, log cos B = log cot h -\- log tan p. (215) To find h. Make the complement of h the middle part, and we have, by Napier's rules, or by (198), (210) (217) Here, as in the preceding article, h and B are determmcd without ambiguity, for there is only one angle less than 180° cor- responding to a given cosine ; and A must be of the same species as./>. cos h = i COS^ ( =os b; whence. cos b COS h COS !>'' or, by logarithms. log cos b : = log cos h — log cos p- 90 TRIGONOMETRY. EXAMPLES. 1. Given in a right-angled spherical triangle ABC, the hv- pothenuse h equal to 91° 42', and the side jo equal to 95° 22 30"; to solve the triangle. Ans. A, 95° 6'; B, 71° 36' 45"; h, 71° 32' 12". 2. Given in a right-angled spherical triangle, the hypothenuse equal to 70° 23' and a side equal to 48° 24' ; to solve the tri- angle. Case HI. 174. Griven one side and its adjacent oblique angle. Let there be given (Fig. Art. 172) the side b and the angle A ; to solve the triangle. To find B. Make the complement of B the middle part, and we have, by Napier's rules, or by (207), cos 5 = sin ^ cos b, or, by logarithms, log cos i?= log sin A -f- log cos h. (218) To find p. Make h the middle part, and we have, by Napier's rules, or by (205), sin h = cot A tan p ; whence, tnn^ = tanj4 sin J, (219) or, by logarithms, log tan jB = log tan A -(- log sin b. (220) To find h. Make the complement of^ the middle part, and we have, by Napier's rules, or by (202), cos A =: cot h tan b ; whence, cot A = cos ^ cot b, (221) or, by logarithms, log cot % = log cos A -\- log cot h. (222) BOOK V. 91 KXAMPLES. 1. Given in a spherical triangle ABO, right-angled at C, the side b equal to 29° 46' 8", and the angle A equal to 137° 24' 21"; to solve the triangle. Ans. £, 54° 1' 16"; p, 155° 27' 54" ; h, 142° 9' 13"- 2. Given in a spherical triangle ABC, right-angled at C, the side p equal to 149° 47' 23", and the angle B equal to 80° 40' ; to find the other parts. Case IV. 175. Given one side and its opposite oblique angle. Let there be given in a spherical tri- angle ABC, right-angled at G, the side p and the opposite angle A ; to solve the triangle. To find h. Make p the middle part, and we have, by Napier's rules, or by (200), sin^ = sin A %m h ■. sin h : ' sin ^-i ' log sin p — - log sin A. (223) (224) whence, or, by logarithms, log sin h : To find h. Make h the middle part, and we have, by Napier's rules, or by (205), sin b = cot A tan p, or, by logarithms, log sin b =z log cot A -\- log tan p. (225) To find B. Make the complement oi A the middle part, and we have, by Napier's rules, or by (206), cos A^=.sm B cos p ; cos A cos p whence, sin B = or, by logarithms, log sin B = log coj A — log cos p. (226) (227) 92 TRIGONOMETRY. Here, since all the unknown parts are determined by their sines, and since there are always two angles less than 180° cor- responding to a given sine, h, h, and B may be taken either acute or obtuse ; hence there may be two solutions. For, produce A B and A G till they meet in A!, then we have a second triangle, A' B G, which satisfies the given conditions, ibr it has a right angle at G, the given side/;, and A' equal to A, the given angle. But h', b', and B, the otlier parts of the second tri- angle, are respectively the supplements of A, b, and i? of the first triangle. When, however, p is given equal to A, we have A, b, and B, each equal to 90°, and the triangle A'B G is equal to the tri- angle ABG (Geora., Prop. XII. Bk. IX). When p and A are both equal to 90°, A is also equal to 90°, and b and B are equal, but indeterminate. EXAMPLES. 1. Given in a spherical triangle ABG, right-angled at C, the side p equal to 36° 31', and the angle A equal to 87° 25' ; to solve the triangle. Ans. h, 78° 20', or 101° 40'; h, 75° 26', or 104° 34'; B, 81° 12', or 98° 48', when carried only to minutes. 2. Given in a spherical triangle ABG, right-angled at G, the side h equal to 79° 30', and the angle B equal to 89° 35' ; to solve the triangle. Case V. 176. Given the two sides about the right angle. Let there be given (Fig. Art. 172) the sides p and b; to solve the triangle. To find h. Make the complement of h the middle part, and we have, by Napier's rules, or by (198), cos A :^ cos p cos b, or, by logarithms, log co-i h = log cos p -f- log cos b. (228) BOOK V. 93 To find A. Make b the middle part, and we have, by Na- pier's rules, or by (5^05), sin h = cot A tan p ; whence, cot A = cot ^ sin b, (229) or, by logarithms, log cot A = log cot p -\- log sin I. (2-30) To find B. Make p the middle part, and we have, by Na- pier's rules, or by (204), sin ^ ^ cot 5 tan h ; whence cot B = t~inp cot b, (231) or, by logarithms, log cot B = log sin p -\- log cot b. (232) These formulae determine h, A, and B without ambiguity. EXAMPLES. 1. In a spherical triangle ABC are given the sides about the right angle, p equal to 48° 24' 16", and b equal to 59° 38' 27": to solve the triangle. Ans. h, 70° 23' 42"; A, 52° 32' 55"; B, 66° 20' 40"- 2. Given in a right-angled spherical triangle, the side p equal to 95°22'30", and the side b equal to 71° 32' 14"; to find the other parts. Case VI. 177. Given the two oblique angles. Let there be given (Fig. Art. 172) the angles A and B ; to sohe the triangle. To find h. Make the complement of h the middle part, and we have, by Napier's rules, or by (199), cos h =^ cot A cot B, or, by logarithms, log cos k = log cot A -\- log cot B. (233) To find p. Make the complement of A the middle part, and we have, by Napier's rules, or by (206), 9 94 TRIGONOMETRY. whence, or, by logarithms, log cos p cos ^ = sin i? cos p ; cos A ■^ sm ii (234) log cos A — log sin S. (235) To find h. Make the complement of B the middle part, and we have, by Napier's rules, or by (207), cos B = sin A cos h ; cos B A' (236) (237) whence, cos h = or, by logarithms, log cos h = log cos B ■ — log sin A. Here /(, p, and h are determined without ambiguity. EXAMPLES. 1. In a right-angled spherical triangle ABC are given the two oblique angles, Jl equal to 44°50', and £ equal to C5°49'53"; to solve the triangle. Ans. h, 63° 10' 4" ; p, 38° 59' 11" ; 6, 54° 30'. 2. Given, in a right-angled spherical triangle, the two oblique angles, A equal to 125° 30', and B equal to 80° 40' ; to find the other parts. QUADKANTAL TrIANGLKS. 178. A yuADRANTAL TRiAMjLE is a Spherical triangle hav- ing one of its sides quadrantal, or equal to 90°. Quadrantal triangles may be solved in the same manner as right-angled spherical triangles, by means of the polar triangle. Let A BG be a quadrantal tri- angle, and A'B'O' denote a tri- angle polar to it ; then, by Art. 150, we have ^' = 180°— jt), B': p'=zl80° — A, V : 180°— 5, 180° — B, BOOK V. ' 95 Now, if the side h be taken equal to 90°, its corresponding polar angle C will also equal 90°; hence the polar triangle will be right-angled, and can be solved by application of the preced- ing formuliie for right-angled spherical triangles, and thus the required parts of the quadrantal triangle may be determined. A triangle, one of whose sides is a quadrant, may also be solved by la3ing off a quadrant on one of the other sides, pro- longed if necessary, and connecting this last point with the other extremity of the original quadrant by the arc of a great circle, thus making the original quadrantal triangle either the difference or the sura of a bi-quadrantal and a right-angled spherical tri- angle. Solving the latter solves the original triangle. Thu?, AG" measures B, OA G" — 90° — A, G G" = 90° — p, A G G" = 180° — G, and solving the triangle A G G" also solves the triangle ABC EXAMPLES. 1. Let there be given, in a quadrantal triangle ABG, the side h equal to 90°, the angle A equal to 54° 4o', and the angle B equal to 42° 12' ; to find the other parts. By taking the supplements of the gi\en parts, we have in the polar triangle, p' = 120° 17', b' = 137° 48', whence A', B', and h' are determined as in Art. 176, and the supplements of tliese give the required parts of tlie quadrantal triangle. Ans. p, 64° 34' 40"; b, 48°0'16'; G, 11 5° 20' 5". 2. Given two sides of a quadrantal triangle equal to 72° 53' and 51° 4', to find the angle opposite to the quadrantal side. Ans. 104° 24' 21". SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 179. In the solution of oblique-angled spherical triangles, it is sometimes found convenient, especially in removing an ambi- guity, to refer to one or more of the following propositions, of which the first four have been demonstrated in Book IX. of the Geometry. 9G TEIGONOMETEY. I. Any side of a spherical triangh is less than the sum of the other two'. II. The sum of the sides is less than 360°. III. The sum of the angles is greater than 180°. IV. nie greater side is opposite the greater angle, and con- versely. V. Any angle is greater than the difference hetween 180° and the sum of the other two angles. Vi. A side which differs more from 90° than another side, is of the same species as its opposite angle. For, by (150), we have cos a = cos b cos c -(- sin b sin c cos A ; whence, , cos a — • cos b cos c cos A = -. — =— ^ , sm sin c in which the denominator is always positive. Then, if a ditfers more from 90° than b or than c, cos a is numerically greater than cos b, or than cos c, and we have cos a > cos b cos c ; hence, the sign of the numerator, and consequently the sign of cos A, is the same as that of cos a, that is, A and a arc in the same quadrant. VII. An angle which differs more from 90° than another angle, )s of the same species as its opposite side. For, by (153), we have cos ^ ^ sin 5 sin O cos a ■ — ■ cos B cos G ; whence, cos A ~\- cos B cos C cos a = r „- . ,-, , sm li sni C in which, if A differs more from 90° than B, or than C, cos A is numerically greater than cos B, or than cos G, and the sign of cos a is the same as that of cos A, that is, a and A are in the same quadrant. VIII. When the sum of two sides is greater than, equal to, or less than 180°, the sum of the two opposite angles is the same. BOOK V. 97 For, by means of (188), tan ^ (a -\- b) cos ^ {A -\- B) = tan -^ c cos ,} {A — B), in which the second member is always positive, since ^ c and j (^4 — B) are each less than 90°, ?o that the factors of the fir^t member, tan ^ {a -{' h) and co-i i {A -\- B) must have the same .-'ign. Therefore, i {a -\- b) and ^ (A -\- B) are of the same species. 180. In the solution of oblique-angled spherical triangles, there are six cases, the data in them being, respectively, I. Two sides and an angle opposite one of them. II. Two angles and a side opposite one of them. III. Two sides and the included angle. TV. Two angles and the included side. V. The three sides. VI. The three angles. Case I. 181. Given two sides and an angle opposite one of them. Let there be given, in the oblique- ^ angled spherical triangle A B G, the ^ ^ sides a and b, and the angle A ; to solve the triangle. To find B. We have, from (148), sm B =. - — sm A, sm a or, by logarithms, log sin B = log sin b — log sin a -\- log sin A. (238) To find C and c. We have, by Napier's analogies, (186) and (188), cot i (7 = ^^±±3 tan ^ (A + B), ^ cos ^ (a — J) ^ ^ ' ^' , cos i (^1 + -S) , 1 . I 7 ^ or, by logarithms, 98 TRIGONOMETRY. log cot |- (7 == log COS ^ (a-\-b) — log cos -^ (a — h) + log tan 1(^ + 5), (239) log tan ^ c = log cos ^ {A-\-B) — log cos J- (A — B) + log tan i (a + b), (240) which determine J C and J- c, and thence O and c. In this case, since B is found from its sine, it will sometimes admit of two values, the one supplementary to the other. When B has two values, G and c must each have two corresponding values. Whether both values of B are admissible must be de- termined by one of the propositions of Art. 179. Thus (Prop. VI.), if b differs more from 90° than a, B must be of the same species as b, and there can be but one solu- tion ; but if b differs less from 90° than a, there may be two solutions. Or (Prop. VIII.), if only one of the supplementary values of B makes ^ {A-\- B) of the same species as J (ra -)- b), there can be but one solution ; but if both values of B fulfil that con- dition, there will be two solutions. EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the side a equal to 63° 50', the side b equal to 80° 19', and the atigle A equal to 51° 30'; to solve the triangle. Solution. By (238) we have a = 63° 50' ar. co. log sin 0.046958 b = 80° 19' log sin 9.993768 ^ = 51° SO' log sin 9.893544 B = 59° 15' 57", or 120° 44' 3" log sin 9.934270 As b differs less from 90° than a, both values of B are admis- sible, and we have i (b—a) = 8° 14/ 30", 4 (a + 5) = 72° 4' 30", i{A-\-B}=OD° 22' 58" or 86° 7' 2' , and ^ {B—A) = 3° 52' 58" or 34° 37' 2". The cosines of ^ (b—a) and J (B — A) are the same as those of J (a — b) and J- (A — B), respectively, by Art. 79. Hence, by (239) and (240), BOOK V. 99 J {h—a) ar. co. log co^+ 0.004508 ar. co. log cos+ 0.004508 ^(a_J_i) log co;+ 9.4882-29 log cos4- 9.488229 ^(^A-^B) logtan+10.1G()9G4 or log tan-f 1 1.108314 i-G log cot-J- 9.653701 oi- logcot-l-10.6G1051 1(7= 65° 44' 53" or 12° 18' 42", (7= 131° 29' 46' or 24° 37' 24". h{B—A) ar.co.logcos+ 0.000998 or 'ar.co.logco--i- 0.084618 ^{A-\-B) log C0S+ 9.754418 or log oos-f 8.830687 ^(a_|_6) logtan-)-10,49016l log tan-)-10.490161 4-c log tan-1-10.245577 or log tan4- 9.405466 ic= 60° 23' 57" or 14° 16' 18", c= 120° 47' 54" or 28° 32' 36"- 2. Given, in an oblique-angled spherical triangle, two sides equal to 99° 40' 48" and 64° 23' 15", and an angle opposite to the first of these equal to 95° 38' 4" ; to find the other side and angles. . Ans. Side, 100° 49' 30"; angles, 65° 33' 10" and 97° 26' 30". Case II. 182. Given two angles and a side opposite one of them. Let there be given, in the oblique-angled spherical triangle ABC (Fig. Art. 181), the angles A and B, and the side a ; to solve the triangle. To find b. We have, from (148), , sin B . sin = ; 7 sm a, sin A or, by logarithms, log sin h = log sin B — • log sin A -\- log sin a. (241) To find C and c. We use equations (239) and (240), as in the last article. This case is exactly analogous to Case I., and gives rise to the same ambiguities, as may be shown by passing to the polar tri- angle. 100 TRIGONOMETRY. If B (lifFurs more from 90° than A, h must be of the same species as B, and there can be but one solution ; but if B differs less from 90" than A, there may be two solutions. (Prop. VII. Art. 179.) Or, if only one of the supplementary values of h makes ^ {a-\-h) of the same species as ^ {A-\- B), there can be but one solution ; but if both values of h fulfil that condition, there will be tviro solutions. (Prop. VIII. Art. 179.) EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the angle A equal to 135°, the angle B equal to 60°, and the side a equal to 155° ; to find the other parts. Ans. G, 98° 3' 4" or 16° 57' 1"; h, 31° 10' 17" or 148° 49' 43" ; c, 143° 42' 57" or 10° 2' 6". 2. Given, in an oblique-angled spherical triangle, two angles equal to 97° 26' 30" and 65° 33' 10", and the side opposite to the first equal to 100° 49' 30"; to find the other parts. Case III. 183. Given two sides and the included angle. Let there be given, in the oblique-angled spherical triangle AB G (Fig. Art. 181), the sides a and h, and the included angle C; to solve the triangle. To find A and B. By means of Napier's analogies (186) and (187), we have tanH^ + i?)="M'^JJcoti6', " ^ ' ^ cos -^ (n -|- 6) , / J ji\ sin 4 (a — J) ^ , „ tan X (A — -B) = . , ' — r-J^ cot J- G ; ■^ ^ ^ sin ^ (n -)- 0) ■^ or, by logarithms, log tan J {A-\-B) = log cos i (a — b) — • log cos ^ (u-\-l>) + logcotiO, (242) log tan J (A — -B) = log sin i (a — b) — • log sin J (a-\-b) -f log cot J- G, (243) which determine J- (A-\-B) and ^ (A — B). The sum of these values gives A, and the second subtracted from the first gives B. BOOK V. 101 To find c. We use equation (240), as in the first two cases. The value of c might also be obtained hy (147) or (149) ; but as it is thus determined from its sine, it would be necessary to re- move the ambiguity by means of the principles contained in Art. 179. As A, B, and c may all be found by means of tangents, there can be but one value for each. It will be observed that J- {A-\-B) nmst always be of the same species as ^ (a -|- V). (Prop. VIII. Art. 179.) EXAMPLES. 1. Given, in an oblique-angled spherical triangle ABC, the side a equal to 70°, the side b equal to 38° 30', and the included angle C equal to Sl° 34' 26" ; to solve the triangle. Solution. J. {a-\-b) = 54° 15', ^ (a — 5) = 15° 45', and i O =. 15° 47' 13"; then. By (242), By (243), i.(a_^5) ar.co.logcos-l- 0.233402 ar. co. log sin + 0.090672 ?,{a — h) log cos + 9.983381 log sin -f 9.433675 i C log cot -1-10.548635 log cot -(-10.548635 ^ {A -f B) log tan -^10.765418 a (^—5) logtan-^ 10.072982 h{A-{- B) = 80° 15' 41" ^ {A~ B) = 49° 47' 30" A = 130° 3' 11" B = 30° 28' 11" By (240), ^ {A — B) = 49° 47' 30" ar. co. log cos -|- 0.190058 ^ (A-\- B) = 80° 15' 41" log cos -\- 9.228282 ^ (a + b) ;= 54° 15' log tan -\- 10.142730 4 c =20° log tan -|- 9.561070 Ans. Angle A, 130° 3' 11" ; angle B, 30° 28' 11" ; side c, 40°. 2. Given, in an oblique-angled spherical triangle, an angle equal lo 48° 36', and the two adjacent sides equal to 112° 22' 58i" and 89° 1 6' 53|" ; to find the other parts. 102 TRIGONOMETRY. Case IV. 184. Given two angles and the included side. Let there be given, in the obhque-angled spherical triangle ABO (Fig. Art. 181), the angles A and B, and the included side c ; to solve the triangle. To find a and b. By means of Napier's analogies (188) and (189), we have tanH« + ^)= °°i^1lg fcina,, •^ ^ ' ' cos\ {A -\- B) ■^ ' tan 4. (a — o) ^ . -"4 . . ' tan A c ; or, by logarithms, log tan ^ (a -|- i) ^ log cos J- ( J. — B) — log cos J ( J. -}- B) + log tan J. c, (244) log tan J- (a — 5) =log sin i (^ — B) — log sin ^ (A-\-B) 4-logtanic, (245) which determine i {a -\- l>) and ^ (a — b), and thence a and J. To find O. We use equation (239), as in the first two cases; but (147) or (149) may be employed, as in the last case. This case is analogous to Case III., and gives rise to no ambi- guity. EXAMPLES. 1. Given, in an oblique-angled spherical triangle ABO, the angles A and 5 equal to 119° 15' and 70° 39', and the side c equal to 52° 39' 4" ; to solve the triangle. Ans. Sides a and b, 112° 22' 58^" and 89° 16' 53^"; angle O, 48° 3G'. 2. In an oblique-angled spherical triangle, given two angles equal to 130° 3' 11" and 31° 34' 26", and the included side equal to 38° 30' ; to find the other parts. Case V. 185. Given the three sides. Let there be given, in the oblique-angled spherical triangle ABO (Fig. Art. 181), the sides a, b, and c ; to solve the tri- anjjle. * BOOK V. 103 To find A, B, and C, we have, by (164), (165), and (166), sin i A = 'sill (•^ -h) sin (s- -<^) sin b sin c 'sin c-^- -c) sni (s- -a) sin c sin a sin i 5 = W sin (s — a) sin (« — i) sm J^ ^ — »/ .- — .— , sin a sin b sin i (7 = i/^ or, by logarithms, . , , loKsinfs — &)-l-log sin (n — c) — Wsini — lofrsinc ,-,,„. logsin^^ = -^^ 5^ i^^^s -V^-^ >^ ^— - , (-246) lo-sin'5 log sin (s— c)+log sin (s— q) —log sin c— log sin a log si n (s— a)+log sin (.s— ft)— log sin a— log s in b logsinJ-G=— ^^ s • K^^°) A, B, and G can also be determined by formulije (167), (168), and (169) for the cosine of half an angle, and by formulje (170), (171), and (172) for the tangent of half an angle. Since the half-angles must be less than 90°, there is no ambi- guity in determining the angles by any of these fbrmulai. EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the side a equal to 70°, the side h equal to 38°, and the side c equal to 40° ; to find the angles. Solution. s = 74°, « — a = 4°, * — 6=36°, s — c = 34°. By (246), By (247), By (248), s — h, logsin9.769219 log sin 9.769219 s _ c, logsin 9.747562 log sin 9.747562 s — a, log sin 8.843585 log sin 8.843585 b, ar.co.logsin0.210658 ar.co.logsin 0.210658 c, ar.co.logsin0.191933 ar.co.log sin 0.1 9 1933 a, ar.co.logsin 0.027014 ar.co.logsin0.027014 2)19.919372 2) 18.810094 2 ) 18.850476 J A, log sin 9.959686 ^ 5,log sin 9.405047 J C, log sin 9.425238 104 TRIGONOMETRY. J ^ = 65° 41' 33".7 i 5 = 14° 43' 18" i G= 15° 26' 21".7 Ans. J, 131° 23' 7"; 5, 29° 26' 36"; G, 30° 52' 43". 2. Given, in an oblique-angled spherical triangle, the sides equal to 112° 22' 59", 89° 16' 53", and 52° 39' 4" ; to solve the triangle. Case VI. 186. Given the three angles. Let there be given, . in the oblique-angled spherical triangle ABC (Fig. Art. 181), the angles A, B, and G ; to solve the tri- angle. To find a, b, and c, we have, by (175), / — cos S cos (S — sin 4- a = 4/ , — ^—. — ^- ■^ y sin iJ sin C I — • cos S cos (S — -B) sin 4- S ^ / — cos S cos {S — j4) ^ " Y sin C sin ^ / — cos S cos (S — C) sm 4 c = t / . — . -. -„ ; ^ y sm A sin B or, by logarithms, log cos S-l-log cos (S — A) — log sin iJ— log sin C fcy.n\ log sin ^ a ^= ^ ) yiiVd ) , . , , log cos S+log cos (5— -B)— log sin C— log sin A log sin 4- o = -^ ' 2 ' V'^''"^ log cos /S-l-log cos (5 — C) — log sin A — log sin S /oki \ log sin ,3-0 = ^ • (^51; a, h, and c can also be determined by formulae (176) for the cosine of half an angle, and by formulae (177) for the tangent of half an angle. Here a, b, and c are determined without ambiguity. EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the angle A equal to 120° 43' 37", the angle B equal to 109° 55' 42", and the angle C equal to 116° 38' 33"; to find the sides. Ans. a, 115° 13' 26"; b, 98° 21' 39'; c, 109° 50' 20". 2. . Given the angles in an oblique-angled spherical triangle equal to 131° 23' 7", 29° 26' 36", and 30° 52' 43"; to solve the triangle. BOOK VI. APPLICATIONS OF SPHERICAL TRIGONOMETRY TO ASTRONOMY AND GEOGRAPHY. 187. The Celestial Sphere is the spherical concave sur- rounding the earth, in whicli all the heavenly bodies appear to be situated. 188. The Zenith is that pole of the horizon which is directly overhead. 189. The Altitude of a heavenly body is its distance above the horizon, measured on the arc of a great circle passing- through that body and the zenith. 190. The Declinatxox of a heavenly body is its distance north or south of the celestial equator, measured on a meridian. 191. The altitude of the celestial pole is equal to the latitude of the place where the observer is located. For the distance from the zenith to the celestial equator is the latitude of the place, and the distance from the zenith to the pole is its complement ; but the distance from the zenith to the pole is also the complement of the altitude of the pole ; hence the latitude of the place and the altitude of the pole are equal. 192. To find the time of the EIS- ING AND SETTING OP THE SUN at any place, the sun's declination and the latitude of the place being given. Let /* represent the celestial north pole, E A Q the celestial equator, H A the rational horizon, S the place of the sun's rising, S' the posi- 10 106 TRIGONOMETRY. tion of the sun at 6 o'clock, PEP' the meridian of the given place, P P P the meridian pawing through S, and PAP' the meridian 90° di-.tant from PEP', passing through S'. From the time of the sun's rising to 6 o'clock, it will pass over S S , the arc of a small circle, corresponding to B A, (he arc of a great circle. Tlie length of ^^, expressed in time (Art. 147), will then give the amount to be taken from or added to 6 o'clock, to give the time of the sun's rising or setting. B S \% the sun's declination, P is the latitude of the place (Art. 191), and Q 0, which measures the angle B A S, is its complement ; hence, in the right-angled spherical triarigle A B S, there are known the side B S and the angle B A S, from which, by Art. 175, sin BA^ tan B S cot B A S, or, log sin BA = log tan sun's decl. -\- log tan lat. of place. After reducing the arc B A to time, at the rate of 15° to an hour, or 4m. to a degree, it must be added to G o'clock for the time of the sun's setting, and subtracted for its rising, wlien the declination and latitude are both north or both south ; but sub- tracted for its setting and added for its rising, when one is north and the other south. The preceding reasoning rests upon the assumption that the sun's declination does not change between sunrise and sunset, which, although not strictly true, is accurate enough for our present purpose. The time obtained is apparent time, and a correction must be applied if we wish to find mean time, or that indicated by the clock. Another correction is necessary for refraction. Neither of these corrections has, however, been applied to the answers that follow. EXAMPLES. 1. Required the time of the sun's rising and setting in Edin- burgh, latitude 55° 57' N., when the sun's declination is 23° 28' N. Ans. Eises, 3h. 20m. 7s. ; sets, 8h. 39m. 53s. 2. "What is the time of the sun's rising and setting in latitude 60° 3' N., when the sun's declination is 23° 28' S. ? Ans. Rises, 9h. lom. 33s. ; sets, 2h. 44m. 27s. BOOK VI. 107 3. Required the time of the sun's rising and setting in places whose latitude is 48° S., when the sun's declination is 15° S. 193. To find the hour of the day at any place, the latitude of the place and the sim's declination and altitude being given. Let Z represent the zenith, and y Z J) an arc of a great circle drawn ^ ^ ro ° TT.X I _i>,l through the zenith and the sun's place, S; P B P, a meridian drawn through the sun's place, &c., as in j_W the last article. As before, the arc A B, added to or subtracted from 6 o'clock, will X^ \, y'Q give the time when the sun is at S; but it will be more convenient to use its complement, E B, which is the time before or after 12 o'clock. E Z\i the latitude of the place, and P Z \% its complement; B S\i the sun's declination, and P S \% its complement ; S D if, the altitude of the sun, and Z S\& its complement ; hence, in the spherical triangle Z P S, the three sides are known, and the an- gle Z P S, OT the arc JS B, may be found by Art. 185. If either the sun's declination, or the latitude of the place, is south, it must be considered negative in taking its complement, unless the south pole is taken as a vertex of the triangle, when north will be negative. EXAMPLES. 1. Required the apparent time of day in the morning, at a place in latitude 39° 54' N., the sun's declination being 17° 29' N., and its corrected altitude 16° 54'. Ans. 6h. 25m. 30s. A. M. 2. In latitude 36° 39' S., when the sun's declination was 9° 27' N., its corrected altitude was observed, in the afternoon, to be 10° 40' ; what was the apparent solar time ? Ans. 4h. 36 m. 10s. P. M. 3. Required the apparent time of day in Boston, latitude 42° 21' N., when the sun's declination is 20° S., and its corrected altitude 15° 15', the sun being east of the meridian. 108 TRIGONOMETRY. 194. To find the shortest distance between two places on the earth's surface, and the bearing of one from the other, their latitudes and longitudes being given. Let Z and S represent the two points on the earth's surface, and P the. north pole of the earth. P Z and P S are the complements of the lati- tudes of the two places, and the arc EB, or the angle ZP S, is the differ- ence of their longitudes ; hence, in the spherical triangle »S' P Z, the two sides P Z and P S, and their included an- gle P, are known, from which the side Z S, and the angles Z S P and S Z Pmstj be found by Art. 183. The distance Z S can easily be reduced to miles by allowing 69.16 statute miles, or 60 nautical miles, to a degree. The an- swers which follow are given in statute miles. If one place is south and the other north of the equator, the south latitude must be considered negative in taking its comple- ment. examples. 1. What is the distance and bearing of Jerusalem, lat. 31° 47' N., long. 35° 20' E., from London, lat. 51° 30' N., long. 6' W. ? Ans. Distance, 2248 miles; bearing, S. 66° 31' E. 2. Required the distance and bearing of Cape Horn, lat. 55° 58' S., long. 67° 21' W., from London. Ans. Distance, 8363 miles; bearing, S. 36° 59' ^V. 3. Required the distance and bearing of (!Julto, lat. 0°, long. 78° 45' W., from San Francisco, lat. 37° 49' N., long. 122° 14' W. TABLE, CONTAINISG THE LOGARITHMS OF NUMBERS FEOM 1 TO 10,000. No. Log. No. Log. No. Log. No. Log. No. Log. 1 0.000000 21 1.322219 41 1.612784 61 1.785330 81 1.908485 2 0.301030 22 1.342423 42 1.623249 62 1.792392 82 1.913814 3 0.477121 23 1.361728 43 1.633468 63 1.799341 83 1.919078 i 0.602060 24 1.380211 44 1.643453 64 1.806180 84 1.924279 o 6 0.698970 25 1.397940 45 1.653213 65 1.812913 85 1.929419 0.778151 26 1.414973 46 1.602758 66 1.819544 80 1.934498 7 0.845098 27 1.4313B4 47 1.672098 67 1.82G075 87 1.939519 8 0.903090 28 1.447158 48 1.681241 68 1.832509 88 1.944483 9 0.954243 29 1.462398 49 1.690196 69 1.838849 89 1.949390 10 1.000000 30 1.477121 50 1.698970 70 1.845098 90 1.954243 11 1.041393 31 1.491302 51 1.707570 71 1.851258 91 1.959041 12 1.079181 32 1.505150 52 1.710003 72 1.857332 92 1.9C3788 13 1.113943 33 1.518514 ■53 1.724276 73 1.863323 93 1.968483 14 1.146128 34 1.631479 54 1.732394 74 1.869232 94 1.973128 15 1.176091 35 1.544068 55 1.740363 75 1.875061 95 1.977724 16 1.204120 36 1.556303 56 1.748188 76 1.880814 96 1.982271 17 1.230449 37 1.568202 67 1.755875 77 1.886491 97 1.986772 18 1.255273 38 1.579784 58 1.763428 78 1.892095 98 1.991226 19 1.278754 39 1.591065 59 1.770862 79 1.897U27 99 1.995636 20 1.301030 40 1.602060 60 1.778151 80 1.903090 100 2.000000 2 LOGARITHMS N. 1 2 3 V 1 5 6 7 1 8 j 9 D. 432 100 000000 000434 000868 U01301 )017341 002166 002598 003029 003461 0038911 1 4321 4751 5181 5609 6038 6466 6894 7321 7748 8174 428 •I 8U00 9026 9451 9876 010300 010724 011147 011570 011993 012415 424 3 012837 013259 013680 014100 4521 4940 5360 6779 6197 6616 420 4 7033 7451 7868 8284 87Ci;, 9116 9532 9947 020361 020776 416 6 021189 021603 022016 022428 02284 !| 023262 023664 024075 4486 4896 412 6 5300 6715 6125 6533 6942 7350 7757 8164 8571 8978 408 1 9384 978,9 030195 030600 031004 031408 031812 032216 032619 033021 404 8 033424 033826 4227 4628 6029 5430 6830 6230 6629 7028 400 9 7426 7825 8223 8620 9017 9414 0811 040207 040602 040998 397 110 041393 041787 042182 042576 042969 043302 013755 044148 044540 044932 393 1 5323 5714 6105 6495 6885 7275 7664 8053 8442 8830 390 2 9218 9G06 9993 050380 050"(:6 f'>U53 051538 051924 062309 052694 386 3 053078 053463 053846 4230 4613 4996 5378 6760 6142 6624 383 4 6905 7286 7666 8046 8426 8805 9185 9563 9942 060320 373 5 060 (i98 061076 061452 061829 062206 062582 062958 063333 063709 4083 376 6 4458 4832 5206 6680 6953 6326 6699 7071 7443 7816 373 7 8186 8557 8928 9298 9668 070038 070407 070776 071145 071514 370 8 071882 072250 072617 072986 073352 3718 4085 4451 4816 6182 366 91 5647 5912 6276 6640 7004 7368 7731 8094 8457 8819 363 120 079181 079543 079904 080266 080626 080987 081347 081707 082067 082426 360 1 082785 083144 083503 3861 4219 4576 4934 6291 6647 6004 357 2 6360 6716 7071 7426 7781 8136 8490 8845 9198 9552 355 3 9905 090258 090611 090963 091315 091667 092018 092370 092721 093071 352 4 093422 3772 4122 4471 4820 5169 5518 5866 6215 £562 349 5 6910 7257 7604 7951 8298 8644 8990 9335 9681 100026 346 6 100371 100716 1010591101403 101747 102091 102434 102777 103119 3462 343 7 3804 4146 44871 4828 6169 5510 6861 6191 6531 6871 341 8 7210 7549 78881 8227' 8505 8903 9241 9579 9916 110253 338 9 110590J11092G 111263;iU599illl934 112270 112605 112940 113275 3609 336 130 113943 114277 1U61 11149441115278 115611 115943 116276 116608 116940 333 1 7271 76031 7934 8265! 8595 8926 9266 9586 9915 120245 330 2 120574 1209031121231, 12160Oil21888 122216 122544 122871 123198 3525 328 3 3852 4178 4504 4830; 5156 5481 6806 6131 6456 6781 325 4 7105 7429 7753 8076] 8399 8722 9045 9368 9690 130012 323 5 130334 130655 130977 131298,131619 131939 132260 132680 132900 3219 321 6 3539 3858! 4177 4496 4814 5133 6451 6769 6086 6403 318 7 6721 7037 7354 7671 7987 8303 8618 8934 9249 9564 316 8 9879 140194 140508 140822 141136 141450 141763 14207C 142389 142702 314 9 1430151 3327 3639 3951 4263 4574 4885 6196 6607 6818 311 140:146128 146438 1467481147058 147367 147676 147985 1 48294 148603 148911 309 1 9219 9527 9835il50142 150449 ' 50756 151063 151370 151676 151982 307 2 152288 152594 152900 3205 3510 3815 4120 4424 4728 5032 306 3 5336 5040 6943 6246 6649 3852 7154 7467 7759 8061 303 4 8362 8664 8965 9266 9567 9868 160168 160469 160769 161068 301 5 161368 161667 161967 162266 162564 162863 3161 3460 3768 4056 299 6 4353 4650 4947 5244 6541 5838 6134 6430 6726 7022 297 7 7317 7613 7908 8203 8497 8792 9086 9380 9674 9968 295 8 170262 170555 170848 171141 171434 171726 172019 172311 172603 172895 293 9 3186 3478 3769 4060 4351 4641 4932 6222 5512 6802 291 150 176091 176381 176070 176959 177248 177536 177825 178U3 IT8T0T 178689 289 1 8977 9264 9552 9839 180126 180413 180699 180986 181272 181558i287 2 181844;i82129 182415il82700 2985 3270 3555 3839 4123 4407;285 3 4691 4975 5259 6542 6825 6108 6391 6674 6956 7239 283 4 7521 7803 8084 8366 864" 8928 9209 9490 9771 190051 281 6 190332 190612 190892 191171 191451 191730 102010 192289 192667 2846 279 b 3125 3403 3681 3959 4237 4514 4792 5069 5346 0623 278 7 5900 6176 6453 6729 7005 7281 7556 7832 8107 8382 276 8 8657 8932 9206 948i; 9756 200029 200303 200577 200850 201124 274 9 201397 201670 201943'202216:202488 2761 3033 3305 3577 3848 272 r 1 1 2 1 3 1 4 II 5 6 1 7 « 1.,?- ■5: or NUMBERS. N. 1 1 2 1 3 1 4 1 5 1 6 7' i" 8■■ 9 iii. ieo 204 12U 204391 2U46i;31 204934 205204 205475 205740 20601620U286 2066661271 1 (5820 70;i6 7365' 7634 7904 8173 8441 8710 8979 92471269 2 9515 9783 210051 210319 210586 210853 211121 211388 211654 2119211267 3 212188 212454 2720 2986 3252 3518 3783 4049 4314 4579:266 4 4844 5109 5373 5638 5902 6166 6430 6694 6957 7221 204 6 7484 7747 8010 8273 8330 8798 9060 9323 9585 9846 262 6 220108 220370 220031 220892 221153 221414 221675 221936 222196 222450 261 7 2716 2976 3236 3496 3755 4015 4274 4533 4792 6051 259 8 5309 5568 5826 6084 6342 6600 6858 7115 7372 763012581 9 7887 8144 8400 8657 8913 231470 9170 9426 9682 9938!230193|256| 170 230449 230704 230960 231215 "231724 231979 232234 232488 232742,256! 1 2996S 3250, 3504, 3757 4011 4264 4517 4770 5023 5276 263 2 5528 - 8046 5781 6033 6285 6537 6789 7041 7292 7544 7795 252 3 8297 8548 8799 9049 9299 9550 9800 240050 240300 250 4 240549 240799 241048 241297 241546 241795 242044 242293 2541 2790 249 6 3038 3286 3534 3782 4030 4277 4525 4772 6019 6266 248 6 6513 5759 6006 6252 6499 6745 6991 7237 7482 7728 240 7 7973 8219 8464 8709 8954 9198 9443 9687 9932 250176 246 8 250420 250664 250908 251151 251395 251638 261881 252125 252368 2610 243 9 2853 3090 3338 255755" 3580 "255996 3822 256237 4064 256477 4306 4543 4790 5031 242 180 255273 255514 256718 256958 257198 257439 241 1 7679 7918 8158 8398 8637 8877 9116 9355 9594 9833 239 2 260071 260310 260548 260787 261025 261263 261501 261739 261976 262214 238 3 2451 2088 2925 3162 3399 3636 3873 4109 4346 4582 237 4 4818 5054 5290 5525 5761 6996 6232 6467 6702 6937 235 6 7172 7406 7641 7875 8110 8344 8578 8812 9046 9279 234 6 9513 9746 9980 270213 270446 270679 270912 271144 271377 271609 233 7 271842 272074 272306 2538 2770 3001 3233 3464 3696 3927 232 8 4158 4389 4620 4850 5081 6311 6542 6772 6002 6232 230 9 6462 6692 278982 6921 279211 7151 279439 7380 7609 7838 8067 8296 8525 22U 1901278754 279667 279895 280123 280351 280578 280806 228 11281033 281661 281488 281715 281942 282169 2396 2622 2840 3075 227 2 3301 3527 3753 3979 4205 4431 4650 4882 6107 6332 226 S 5557 5782 6007 6232 6456 6681 6905 7130 7354 7578 225 4 7802 8026 8249 8473 8696 8920 9143 9366 9589 9812 223 5 290035 290257 290480 290702 290925 291147 291369 291591 291813 292034 2221 6 2256 2478 2699 2920 3141 3363 3584 3804 4025 4246 221 7 4466 4687 4907 5127 5347 6567 6787 6007 6226 6446 220 8 6665 6884 7104 7323 7542 7761 7979 8198 8416 8635 219 9 8853 9071 9289 9507 9725 9943 300161 300378 300595 300813 218 200 301030 301247 301464 301681 301898 302114 302331 302547 302764 302980 Wl 1 3196 3412 3628 3844 4059 4275 4491 4706 4921 5136 216 2 6351 5566 5781 5996 6211 6425 6639 6854 7068 7282 215 3 7496 7710 7924 8137 8351 8564 8778 8991 9204 9417 213 4 9630 9843 310056 310268 310481 310693 310906I311118 311330 311542 212 6 311754 311966 2177 2389 2600 2812 3023 3234 3445 3666 211 6 3867 4078 4289 4499 4710 4920 5130 6340 5551 5760 210 7 5970 6180 6390 6599 6809 7018 7227 7436 7646 7854 209 8 8063 8272 8481 8C89 8898 9106 9314 9522 9730 9938 208 9,320141; 320354 320o62|320769 3209771321184 321391 321598 3'23"665 321805 322012 207 210 322219 322426 322633 322839 323046 323252 323458 323871 324077 206 1 4282 4488 4694 4899 6105 5310 6516 6721 . 5926 6131 205 2 6336 6541 6745 6950 7155 7359 7563 7767 7972 8176 204 3 8380 8583 8787 8991 9194 9398 9601 9805 330008 330211 203 4 330414 330617 330819 3?i022 331225 331427 331030 331832 2034 2236 202 & 2438 2640 2842 3044 3246 3447 3649 3850 4051 4253 202 6 4454 4655 4856 5057 6257 6458 5658 5859 6059 6260(201 7 6460 6660 6860 7060 7260 7459 7059 7858 8058 8257 200 8 8456 8656 8855 9054 9253 9451 9650 9849!340047 340246 199 9 340444 3406421340841 341039i341237 341435 341C32'341830i 2028 22251138 N. 1 1,,.^ 3 1 4 , 5 U 1 7 1 8 9 ,D. L0RA.R1THMS N. 1 2 3 4 II 5 6 1 7 1 8 9 D 220 342423 342620 342817 343014 343212 343409 343606 343802 343999 344196 197 1 4392 4589 4785 4981 6178 5374 6570 5760 6962 6167 196 2 6353 6549 6744 6939 7135 7330 7525 7720 7915 8110 195 3 8305 8500 8094 8889 9083 9278 9472 9606 9860 350064 194 i 350248 350442 350U36 350829 361023 361216 351410 361(03 361796 1989 193 6 2183 2375 2568 2761 2954 3147 3339 3532 3724 3916 193 6 4108 4301 4493 4685 4876 5068 5260 5452 5643 5834 192 7 0026 6217 6408 6599 6790 6981 7172 7363 7654 7744 191 8 7935 8125 8316 8506 8696 8886 9076 9266 9456 9646 190 9 9835 360025 3602151360404 360593 360783 360972 361161 361360 361539 189 230 361728 361917 362105 362294 362482 362671 302859 363048 363236 363424 188 1 3612 3800 3988 4176 4363 4551 4739 6610 4926 5113 6301 188 2 6488 5675 6862 6049 6236 6423 6796 6983 7169 187 3 7356 7542 7729 7915 8101 8287 8473 8659 8845 9030 186 4 9216 9401 9587 9772 9958 370143 370328 370513 370698 370883 185 5 371068 371263 371437 371622 371800 1991 2175 2360 2544 2728 184 6 2912 3096 3280 3464 3647 3831 4016 4198 4382 4565 184 7 4748 4932 5115 5298 5481 6664 6846 6029 6212 6394 183 8 6577 6759 6942 7124 7306 7488 7670 7862 8034 8216 182 9 8398 8580 8761 8943 9124 9306 9487 9668 9849 380030 181 240 380211 380392 380573 380754 380934 381115 381296 381476 381656 381837 I8l 1 2017 2197 2377 2557 2737 2917 3097 3277 3466 3636 180 2 3815 3995 4174 4353 4533 4712 4891 5070 5249 6428 179 3 5606 5785 6964 6142 6321 6499 6677 6856 7034 7212 178 4 7390 7568 7746 7923 8101 8279 8456 8634 8811 8989 178 5 9166 9343 9520 9698 9875 30OO51 390228 390405 390582 390759 177 C 390935 391112 391288 391464 391641 1817 1993 2169 2345 2521 176 7 2697 2873 3048 3224 3400 3575 3751 3926 4101 4277 176 8 4452 4627 4802 4977 5152 6326 6501 6676 5850 0025 175 9 6199 6374 6548 6722 6896 7071 7245 7419 7592 7766 174 250 397940 398114 398287 398461 308634 398808 398981 399164 399328 399501 173 1 9674 9847 400020 400192 400305 400538 400711 400883 401056 401228 173 2 401401 401573 1745 1917 2089 2261 2433 2606 2777 2949 172 3 3121 32D2 3464 3636 3807 3978 4149 4320 4492 4663 171 4 4834 5005 6176 5346 5517 5688 6868 0029 6199 6370 171 6 6540 6710 6881 7051 72i,l 7391 7561 7731 7901 8070 170 6 8240 8410 8579 8749 8918 9087 9257 9426 9595 9764 169 7 9933 410102 410271 410440 410609 410777 410946 411114 411283 411461 169 8 411620 1788 1956 2124 2293 2461 2629 2796 2964 3132 168 9 3300 3467 3635 3803 3970 4137 4305 4472 4639 4806 167 260 414973 415140 415307 416474 415641 415808 415974 416141 416308 416474 167 1 6641 6807 6973 7139 7306 7472 7638 7804 7970 8135 166 2 8301 8467 8633 8798 8964 9129 9295 9460 9625 9791 166 3 9966 420121 420286 420451 420616 420781 420945 421110 421275 421439 165 4 421604 1768 1933 2097 2261 2426 2590 2764 2918 3082 164 5 3246 3410 3574 3737 3901 4065 4228 4392 4666 4718 164 C 4882 5045 5208 5371 5534 5697 5860 6023 6186 6349 163 7 6511 6674 6836 6999 7101 7324 7486 7648 7811 7973 162 8 8135 8297 8459 8621 8783 8944 9106 9208 9429 9591 162 9 9762 9914 430075 430236 430398 430559 430720 430881 431042 431203 161 270 431364 431525 431686 431846 432007 432167 432328 432488 432649 432809 161 1 2969 3130 3290 3-150 3610 3770 3930 4090 4249 4409 160 2 4669 4729 4888 6048 5207 5367 6526 6685 6844 6004 159 8 6163 6322 • 6481 6640 6799 6957 7116 7275 7433 7692 159 4 7751 7909 8067 8226 8384 8542 8701 8859 9017 9176 168 5 9333 9491 9648 9806 9964 440122 440279 440437 440594 440752 158 6 440909 441066 411224 4413H1 441538 1695 1852 2009 2166 2323 157 7 2480 2637 2793 2950 3106 3263 3419 3576 3732 3889 157 8 4045 4201 4357 4513 4669 4825 4981 6137 6293 5449 166 9; 6604 6760 5915 6071 6226 6382 6537 6692 6848 7003 166 H 1 1 2 3 4 6 1 6 J 7 1 8 ■ 9 in.j OF NUMBERS. K.i 1 1 2 1 3 1 4 II 5 6 7 a •^'-yix '^80 447158 447313 44746814476231447778 447a33 9478 448088 448242 4483y"7 4485521155 1 8706 8861 9015 9170 9324 9633 9787 9941 450095 154 2 450249 450403 450557 450711 450865 451018 461172 461326 451479 1633 154 3 1786 1940 2093 2247 2400 2553 2706 2859 3012 3165 153 4 3318 3471 3624 3777 3930 4082 4236 4387 4540 4692 153 6 4845 4997 5150 5302 5454 6606 6768 6910 6062 6214 152 6 6366 6518 6670 6821 6973 7125 7276 7428 7579 7731 152 7 7882 8033 8184 8336 8487 8638 8789 8940 9091 9242 151 8 9392 9543 9694 9845 9995 1G0146I460296 460447 460597 460748 151 9 460898;401048 461198 461348 461499 1649' 1799 1948 2098 2248 lo'J 290 462398 462548 462697 462847 462997 463146 463296 463446 463i94 463744 150| 1 3893 4042 4191 4340 4490 4G39 4788 4936 5085 5234 149 2 6383 5532 5680 5829 6977 6126 6274 6423 6571 6719 149 3 6868 701U 7164 7312 7460 7608 7766 7904 8052 8200 14B 1 8347 8495 8643 8790 8938 9086 9233 •JJ80 9527; 9676 148 5 9822 9969 470116 470263 470410 470557 470704 470851 470998471145 147 6 471292 471438 1585 1732 1878 2025 2171 2318, 2404 2610 146 7 2756 2903 30-19 3195 3341 3487 3633 3779 3925 4071 116 8 4216 4362 45US 4653 4799 4944 6090 6235 5381 5526 146 9 5671 5816 5962 6107 6252 6397 6542 6687 6832 6976 145 300 477121 477266 477411 477555 477700 477844 477989 478133 478278 478422 145| 1 8566 8711 8855 8999 9143 9287 9431 9575 9719 9863 144 2 480007 480151 480204 480438 480582 480725 480869 481012 481156 481299 144 3 1443 1586 1729 1872 2016 215H 2302 2445 2588 2731 143 4 2874 3016 3159 3302 3445 3587 3730 3872 4015 4157 143 6 4300 4442 4585 4727 4869 5011 5153 5295 5437 5579 U2 6 5721 5863 6005 6147 6289 6430 6572 6714 6856 6997 142 7 7138 7280 7421 7563 7704 7845 7986 8127 8269 8410 141 3 8551 8692 8833 8974 9114 9255 9396 9537 9677 9818 141 9 9958 490099 490239 490380 4905 2U 490661 490801 490941 491081 491222 140 310 491362 491502 491642 491782 491922 4U2062 492201 4l)234l 492481 492621,140 1 2760 2900 3040 3179 3319 3458 3597 3737 3876 4015'l39 2 4155 4294 4433 4572 4711 4850 4989 6128 6267 5406 139 3 5544 6683 5822 5960 6099 6238 6376 6515 6653 6791 139 4 6930 7068 7206 7344 7483 7U21 7759 7897 8035 8173 138 5 8311 8448 8586 8724 8862 8999 9137 9275 9412 9550ll38 6 9687 9824 9962 500099 500236 600374 500611 500048 500785 500922ll37 7 501059 501196 501333 1470 1607 1744 1880 2017 2154 2291 ll37 8 2427 2564 2700 2837 2973 3109 3246 3382 3518 3655136 9 3791 3927 4063 4199 4336 4471 4607 4743 4878 5014|136 320 505150 505286 505421 505557 505693 605828 605964 5000991506234 506370 136 1 6505 6040 6776 6911 7046 7181 7316 7451 7586 7721 135 2 7856 7991 8126 8260 8395 8530 8664 8799 8934 9068 135 3 9203 9337 9471 9606 9740 9874 510009 610143 510277 610411 134 4 510545 510679 510813 610947 511081 511215 1349 1482 1616 1750134 6 1883 2017 2151 2284 2418 2551 2684 2818 2961 3084133 6 3218 3351 3484 3617 3750 3883 4016 4149 4282 4416'133 7 4548 4681 4813 4946 5079 6211 5344 6476 5609 574l'l33l 8 5874 6006 6139 6271 6403 6635 6668 6800 6932 7064 132 9 7196 7328 7460 7592 7724 7855 7987 8119 8251 8382 132J 330 518514 518646 518777 518909 519040 519171 519303 519434 519566 519697 131 1 9828 9959 5200UO 520221 520353 520484 520615 520746 620876 621007 131 2 521138 521269 1400 1530 1661 1792 la22 2053 2183 2314 131 3 2444 2575 2705 2835 2986 3096 3226 3356 3486 1616 130 4 3746 3876 4006 4136 4266 4396 4620 4656 4785 4915 130 5 5045 6174 5304 5434 5563 5693 5822 6951 6081 6210 129 6 6339 6469 (1598 6727 6856 6985 7114 7243 7372 7501 129 7 7630 775!) 7888 8016 8145 8274 8402 8631 8660 8788 129 8 8917 9045' 3174' 9302; 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OP NUMBLRS. 13 N. 0^ 1 2 3 4 5 « 7 8 9 D. 7B0 880814 880871 880928 880986 881042 881099 881156 881213 881271 881328 67 1 1383 1442 1499 1556 1613 1670 1727 1784 1841 1898 57 2 1955 2012 2069 2126 2183 2240 2297 2354 2411 2468 57 3 2525 2581 2638 2695 2752 2809 2866 2923 2980 3037 67 4 3093 3150 3207 3264 3321 3377 3434 3491 3548 3605 6V 6 3661 3718 3775 3832 3888 3945 4002 4959 4115 4172 57 6 4229 4285 4342 4399 4455 4512 4669 4625 4682 4739 57 7 4795 48S2 4909 4965 5022 5078 6135 6192 5248 5305 57 8 5361 5418 5474 5531 6587 6644 5700 5757 5813 6870 57 9 5926 0383 6039 6096 6162 6209 6265 6321 6378 6434 66 770 886491 886547 886604 886660 886716 886773 8868291886836 886942 886998 "56" 1 7054 7111 7167 7223 7280 7336 7392 7449 7605 7661 66 2 7617 7674 7730 7786 7842 7898 7965 8011 8067 8123 66 3 8179 8236 8292 8348 8404 84G0 8516 8673 8629 8685 66 4 8741 8797 8853 8909 8965 9021 9077 9134 9190 9246 66 5 9302 9358 9414 9470 9526 9582 9638 9694 9750 9806 56 G 9862 9918 9974 890030 890086 890141- 890197 890253 890309 890366 66 7 890421 890477 890533 0589 0646 0700 0766 0812 0868 0924 66 8 0980 1035 1091 1147 1203 1259 1314 1370 1426 1482 66 9 780 1537 1693 1649 1705! 1"60 1816 1872 1928 1983 2039 56 892095 892150 892206 892262 892317 892373 892429 892484 892540 892595 66 1 2651 2707 2762 2818 2873 2929 2985 3040 3096 3161 66 2 3207 3262 3318 3373 3429 3484 3540 3695 3651 3700 66 3 3762 3817 3873 3928 3984 4039 4094 4160 4205 4261 66 4 4316 4371 4427 4482 4538 4693 4648 4704 4769 4814 65 6 4870 4925 4980 5036 6091 5146 5201 6257 6312 6367 55 6 54^3 6478 5533 6688 6644 5699 5754 6809 6864 6920 66 7 6975 6030 6085 6140 6195 6251 6306 6361 6416 6471 66 6 6526 6581 6636 6692 6747 6802 6867 6912 6967 7022 66 9| 7077 7132 7187 7242 7297 7352 7407 7462 7517 7672 65 790 897627 897682 897737 897792 897847 897902 897959 898012 898067 898122 65 1 8176 8231 8286 8341 8396 8451 8506 8561 8615 8670 55 2 8725 8780 8835 8890 8944 8999 9064 9109 9164 9213 65 3 9273 9328 9383 9437 9492 9547 9602 9656 9711 9700 55 4 9821 9675 9930 9985 900039 900094 900149 900203 900268 900312 65 6 900367 900422 900476 900531 0586 0640 0696 0749 0804 0869 65 6 0913 0968 1022 1077 1131 1186 1240 1295 1349 1404 65 7 1458 1513 1567 1622 1676 1731 1786 1840 1894 1948 64 8 2003 2057 2112 2166 2221 2275 2329 2384 2438 2492 64 9 2547 2601 2655 2710 2764 2818 2873 2927 2981 3036 64 800 903090 903144 903199 903253 903307 903361 90341 C 903470 903524 903578 54 1 3633 3687 3741 3795 3849 3904 3968 4012 4066 4120 64 2 4174 4229 4283 4337 4391 4445 4499 4553 4607 4661 54 3 4716 4770 4824 4878 4932 4986 6040 5094 6148 5202 64 4 5256 6310 6364 5418 5472 5526 5580 6634 6688 5742 54 5 5756 6850 6904 5958 6012 0060 6119 6173 6227 6281 54 6 6335 6389 6443 6497 " 6551 6604 6668 6712 6766 6820 54 7 6874 6927 6981 7035 7089 7143 7196 7250 7304 7358 64 8 7411 7465 7519 7573 7626 7680 7734 7787 7841 7895 54 9 7949 8002 8056 8110 8163 8217 8270 8324 8378 8431 64 810 908485 908539 908592 908646 908699 908753 908807 908860 908914 903967 64 1 9021 9074 9128 9181 9235 9289 9342 9396 9449 9503 64 2 9556 9610 9663 9716 9770 9823 9877 9930 9984 910037 53 S 310091 910144 910197 910251 910304 910358 910411 910464 910518 0571 53 4 0624 0678 0731 0784 0838 0891 0944 0998 1051 1104 53 6 1158 1211 1264 1317 1371 1424 1477 1630 1684 1637 53 6 1690 1743 1797 1850 1903 1956 2009 2063 2116 2169 53 1 2222 2275 2328 2381 2436 2488 2541 2594 2647 2700 63 8 2753 2806 2859 2913 2966 3019 3072 3125 3178 3231 63 9 3284 3337 3390 3443 3496 3549 3602 3655 3708 3761 53 BT.I 1 1 2 1 a 4 1 5 6 i 8 9 |D. u LOGAIUTH.MN N.I 1 1 1 2J ._.,. 4 , 1 6 ^ 8 9 |T5.T 820 813814 913807 913920 913973 914026 914079 914132 914184 914237 914290 53 1 4343 4396 4449 4502 4555 4608 4600 4713 4766 4819 53 2 4872 4925 4977 5030 5083 5136 5189 5241 5294 5347 53 3 6400 5453 5505 5558 5011 6604 5716 5709 5822 5875 53 4 5927 5980 6033 6085 6138 6191 0243 6296 6349 6401 53 6 0454 6507 6559 6612 6664 6717 0770 6822 6875 6927 53 C (i980 7033 7085 7138 7190 7243 7295 7348 7400 7453 53 7 7500 7558 7611 7663 7710 7768 7820 7873 7925 7978 52 8 8030 8083 8135 8188 8240 8293 8345 8397 8450 8502 52 9 8555 8607 8659 8712 8704' 8816 8809 8921' 8973 9020I 52 1 B30 919078 919130 919183 919235 919287 919340 919392 9 19444|9I9496|919549 52 1 9001 9653 9706 9758 9810 9802 9914, 9SW;7'920019 920071 52 2 920123 920176 920228 920280 920332 920384 920436 920489 0541 0593 52 3 0045 0697 0749 0801 0853 0906 ■ 0958 1010 1062 1114 52 4 1106 1218 1270 1322 1374 1420 1478 1530 1582 1634 52 6 1G86 1738 1790 1842 1894 1940 1998 2050 2102 2154 52 6 2206 2258 2310 2302 2414 2406 2518 2570 2622 2674 62 7 2725 2777 2829 2881 2933 2985 3037 3089 3140 3192 62 8 3244 3296 3348 3399 3451 3503! 3555 3607 3658 3710 62 9 3702 3814 3865 3917 3909 4021 4072 4124 4176 4228 62 8t0 924279 924331 924383 924434 924486 924538 9/45¥9'i92464r 924693 924744 ■52 1 4796 4848 4899 4951 5003 5054 5106 5157 5209 5261 52 2 5312 5364 5415 6467 5518 5570 5621 5073 5725 5776 52 3 5828 6879 5931 6982 6034 6086 0137 6188 6240 6291 51 4 6342 6394 6445 6497 6548 6600 6651 6702 6754 6805 51 5 6857 6908 6959 7011 7062 7114 7165 7216 7268 7319 61 6 7370 7422 7473 7524 7576 7627 7678 7730 7781 7832 61 71 7883 7935 7986 8037 8088 8140 8191 8242 8293 8345 51 8' 83i)ii 8447 8498 8549 8601 8652 8703 8754 8805 8867 51 9 8908 8959 9010 9061 9112 9163 9215 9266 9317 9368 51 850 929419 929470 929521 929572 929623 929674t929725j929776 929827 929879 51 ] i WM 9981 930032 930083 930134 9301851930230 930287 930338 930389 51 2i'J30440 930491 0542 0532 0643 0694 0745 0796 0847 0898 61 3 OiMa 1000 1051 1102 1153 1204 1254 1305 1350 1407 51 4 1458 1509 1560 1610 1661 1712 1763 1814 1805 1916 61 5 19(10 2017 2008 2118 2169 2220 2271 2322 2372 2423 61 6 2474 2524 2575 2620 2677 2727 2778 2829 2879 2930 61 7 2981 3031 3082 3133 3183 3234 3285 3335 3386 3437 61 8 3487 3538 3589 3039 3690 3740 3791 3841 3892 3943 61 ti 3993 4044 4094 4145 4195 4246 4296 934801 4347 934852 4397 4448 51 860 934498 934549 934599 934650 934700 934751 934902 934963 50 ] 5003 5054 5104 6154 6205 5255 5300 5350 5406 5457 50 2 5507 5558 5608 6658 5709 5759 5809 6860 6910 6960 50 3 con 6061 6111 6162 6212 6262 6313 6363 6413 6463 50 4 6514 6564 6614 6665 6715 6765 6815 6805 6916 6966 50 6] 7ul(i 7066 7117 7167 7217 7267 7317 7367 7418 7468 50 6' 7518 7568 7618 7668 7718 7769 7819 7869 7919 7969 50 7! 8019 8009 8119 8169 8219 8269 8320 8370 8420 8470 50 8 8520 8570 8620 8670 8720 8770 8820 8870 8920 8970 60 9 9020 9070 9120 9170 9220 9270 9320 9309 9419 9469 50 870 039519 9695091939619,939069 939719 939769 939819i939M69i939918 939908 50 1 9400 18, 940008|940118;940l68 940218 940267 940317J940367 940417 940467 60 2 0516 0566 0616 06661 0710 0765 0815 0865 0915 0964 50 3 1014 1004 IIU 1103, 1213 1263 1313 1362 1412 1462 60 4 1511 1561 1611 1660 1710 1760 1809 1859 1909 1958 50 5 2008 2058 2107 2157 2207 2256 2300 2355 2405 2455 50 6 2504 2554 2603 2653 2702 2752 2801 2851 2901 2950 60 7 3000 3049 3099 3148 3198 3247 3297 3346 339( 3445 49 8 3445 3544 3593 3643 3692 3742, 3791 3841 3890| 3930 49 9 3989 4038 4088 4137 4186 4236 4285 4335 43841 4433 49 n] . » ...1 2 3 4 5 j 6 7 8 1 9 |D.| OF NUMBERS. If) iT ■l 2 3 4 5 6 7 8 9 -q B8U 944483 944532 944581 944031 944080 944729 944779 944828 944877 944927 49 1 4970 6025 5074 6124 5173 5222 5272 5321 5370 6419 49 2 5409 6518 5507 5016 5005 5715 5704 5813 5862 5912 49 3 6901 0010 0059 0108 0157 0207 0250 6305 635i 6403 49 4 0452 0501 6551 0000 6049 0698 6747 0790 6845 0894 49 S 0943 6992 7041 7090 7140 7189 7238 7287 7336 7385 49 6 7434 7483 7532 7581 7630 7079 7728 7777 7826 7875 49 7 7924 7973 8022 8070 8119 8108 8217 8206 8315 8364 49 8 8413 8402 8511 8500 8009 8657 8706 8755 8804 8853 49 9 8902 8951 8999 9048 9097 9146 9195 9244 9292 9341 49 §90 949390 949439 9494881949530 949585 949634 949683 949731 949780 949829 49 1 9878 9920 9975 950C24 950073 930121 950170 950219 9502(;7 950310 49 2 950305 950414 950402 0511 0500 0008 0057 0706 0754 0803 49 3 0851 0900 0949 0997 1040 1095 1143 1192 1240 1289 49 i 1338 1380 1435 1483 1532 1580 1029 1077 1726 1775 49 5 1823 1872 1920 1969 2017 2000 2114 2103 2211 2200 48 b 2308 2356 2405 2453 2502 2550 2399 2047 2090 2744 48 7 2792 284) 2889 2D38 2980 3034 3083 3131 3180 3228 48 8 3276 3325 3373 3421 3470 3518 3306 3615 3603 3711 4S 9 3700' 3S08 3850 3905' 3953 4001 4049 4098 4146 4194 48 900954243 9542?! 954339,954387 954435 954484 954532 954580 954028 954677 48 ll 4725 4773 4821 4809 4918 4906 "5014 5062 5110 5158 48 2 6207 5255 5303 5351 6399 5447 5495 6543 5592 6040 48 3 5088 5736 5784 6832 6880 5928 5970 6024 6072 6120 48 4 6108 6210 6265 6313 6361 6409 6457 6505 €553 .6601 48 6 0049 6697 6745 6793 6840 6888 6930 6984 7032 7080 48 6 7128 7170 7224 7272 7320 7308 7410 7464 7612 7559 48 7 7007 7055 7703 7751 7799 7847 7894 7942 7990 8038 48 8 8080 8134 8181 8229 8277 8325 8373 8421 8408 8510 48 9 8504 8012 8059 8707 8755 8803 8850 8898 8946 8994 48 910 959041 959089 959137 959185 959232 959280 959328 959375 959423 959471 48 1 9518 9560 9014 9001 9709 9757 9804 9852 9900 9947 48 2 9995 960042 960090 960138 900185 960233 960281 960328 960370 960423 48 3 900471 0518 0500 0013 006 L 0709 0756 0804 0851 0899 48 4 0940 0994 1041 1089 1136 1184 1231 1279 13 2( 1374 48 & 1421 1409 1510 1503 1611 1038 1706 1753 1801 1848 47 6 1895 1943 1990 2038 2085 2132 2180 2227 2275 2322 47 7 2309 2417 2404 2511 2559 2600 2033 2701 2748 2795 47 8 2843 2890 2937 2985 3032 30^79 3120 3174 3221 3268 47 9 3310 3303 3410 3157 3504 3552 3599 3040 3093 3741 47 920 963788 903835 963882 963929 903977 904024 964071 904118 964165 964212 ll 1 4_00 4307 4354 4401 4448 4495 4542 4590 4037 4084 47 2 4731 4778 4825 4872 4919 4966 5013 6001 5108 5155 47 3 5202 6249 5296 5343 5390 6437 6484 5531 5378 6025 47 i 5072 5719 5700 5813 6800 5907 6954 0001 6048 0095 47 6 0142 6189 0230 0283 6329 6370 6423 6470 6517 6564 47 6 6611 6058 0705 0752 6799 0845 6892 6939 6980 7033 47 7 7080 7127 7173 7220 7267 7314 7361 7408 7454 7501 47 8 7548 7595 7042 7088 7733 7782 7829 7875 7922 7969 47 9 8010 8002 8109 8150 8203 8249 8296 8343 8390 8436 47 930 9684,83 908530 908.)76 968023 908070 908710 968703 908810 9688561 908903 47 1 8950 8990 9043 9090 9130 9183 9229 9270 9323 9369 47 2 9416 9403 9309 9550 9602 9049 9095 9742 9789 9835 47 3 9882 9928 9975 970021 970008 970114 970161 970207 970234 970300 47 4 970347 970393 970440 0483 0533 0379 0020 0072 0719 0705 4L 6 0812 0858 0904 0951 0997: 1044 1090 1137 1183 1229 46 6 1270 1322 1309 1415 li'Jl 1508 1554 1001 1647 1693 46 7 1740 1786 1832 1879 1925 1971 2018 2064 2110 2157 46 8 2203 2249 2295 2342 2388 2434 2481 ■ 2527 2573 2619 46 9 2666 2712 2758 2804 2851 2897 2943 2989 3035 3082 46 N.I 1 1 2 3 4 5 1 6 7 8 9 D. 16 LKGARUHMS OK NUMBERS. N. } 2 1 3 4 1 6 6 7 E 9 |D 940 973128 973174 97;i2-2Uly73266 973313 973359 973405 973451 973497 973543, 46 1 3590 3636 31,82 3728 3774 3820 - 3866 3913 3959 4005 46 2 4051 4097 4143 dl89 4236 4281 4327 4374 4420 4466 46 3 4612 4558 4'J04 1650 4696 4742 4788 4834 4880 4926 46 i 4972 5U18 6U6, 6110 5156 5202 5248 6294 6340 5386 46 5 6432 5478 5524 5570 5616 5662 5707 6753 6799 5845 46 6 5891 5937 6983 6029 6075 6121 6'i67 6212 6258 6304 46 7 6350 6396 6442 6488 6533 6579 6625 6-:71 6717 6763 46 8 6808 6854 6900 6946 6992 7037 7083 7129 7175 7220 46 9 7266 73j2 7358 7403 7449 7495; 7541 7586 7632 7678 46 »>0 977721 977769 977815 977861 977906 977952 977998 978043 978089 97bl35 40 1 8181 8226 8272 8317 8363 8409 8454 8500 8546 8591 46 2 8637 8U83 8728 8774 8819 8865 8911 8956 9002 9047 46 3 9093 9138 9184 9230 9275 9321 9366 9412 9457 9503 46 4 954f 9594 9639 9685 9730 9776 9821 9867 9912 9958 46 6 980003 980049 980094 980140 980185 980231 980276 980322 980367 980412 45 6 0458 0503 0549 0594 0640 0685 0730 0776 0821 0867 U 7 0912 0957 1003 1048 1093 1139 1184 1229 1276 1320 45 8 1360 1411 1456 1501 1547 1592 1637 1683 1728 I7T3 45 9 1819] 18U4i 1909 1954 2000 2045 2090' 2135 2181 2226 45 960 982271 982316 982362 982407 982452 982497 982543 982588 982033 982678 "45 1 2723 2769 2814 2869 2904 2949 2994 3040 3085 3130 45 2 3175 3220 3265 3310 3356 3401 3446 3491 3536 3581 45 3 3626 3671 3716 3762 3807 3862 3897 3942 3987 4032 45 4 4077 4122 4167 4212 4257 4302 4347 4392 4437 4482 45 6 4527 4572 4617 4662 4707 4752 4797 4842 4887 4932 46 6 4977 6022 5067 5112 6157 6202 5247 6292 6337 5382 45 7 5426 5471 6516 5.)6l 6606 5651 6696 6741 5786 5830 45 8 5875 5920 5965 6UI0 6055 6100 6144 6189 6234 6279 45 9 6324 6369 6413 6458 6503 6548' 6593 6637 66S2 0727 45 45 970 986772 986817 986861 986906 986951 986996 987040 987081; 987"! 30 987175 } 7219 7264 7309 7353 7398 7443 7488 7532 7577 7622 45 2 7666 7711 7756 7800 7845 7890 7934 7979 8024 8068 45 3 8113 8157 8202 8247 8291 8336 8381 8425 8470 8514 45 4 8559 8604 8648 8693 8737 8782 8826 8871 8916 8960 45 5 9005 9049 9094 9138 9183 9227 9272 9316 9361 9405 45 6 9450 9494 9539 9583 9628 9672 9717 9761 9806 9850 44 7 9895 9939 9983 990028 990072 990117 990161 990206 990250 990294 44 8 990339 990383 990428 0472i 0516 0561 0605 0660 0694 0738 44 9 0783 0827| 0871 0916 0960 1004 1049 1093 1137 1182 44 980 991226 991270 a91315 991359 991403 991448 991492 991536 991580 991626 "44 \ 1669 1713 1758 1802 1846 1890 1936 1979 2023 2007 44 '2 2111 2156 2200 2244 2288 2333 2377 2421 2405 2509 44 3 2554 2593 2642 2686 2730 2774 2819 2863 2907 2951 44 4 2995 3039 3083 3127 3172 3216 3260 3304 3348| 3391 44 6 3436 3480 3524 3568 3613 3657 3701 3745 3789 3833' 44 6 3877 3921 3965 4009 4053 4097 4141 4185 4229 4 273 44 7 4317 4361 4405 4449 4493 4537 4581 4625 4009 4713i44 8 4757 4801 4845 4889 4933 4977 6021 5065 6108 5152144 9 *)90 5196 995635" 5240! 6284 5328 5372 5416 6460 6504{ 5547 55911 44 9956791995723:995767 995811 9958641996898 995942 995986 990030 44 1 6074 6117 6161 6205' 6249 0293 6337 6380 6424 04C8| 44 2 6512 6555 6599 66431 6U87 7080' 7124 6731 6774 6818 6802 6900 44 3 6949 6993 7037 7168 7212 7255 7299 7343 44 4 7386 7430 7474 7.-. 17 7561 7605 7648 7692' 7736 7779 44 6 7823 7867 7-.)10 79:i4 7998 8041 8085 8I29I 81T2 8216' 44 6 8259 8303 8347 8:!!iO 8434 8477 8521 8564: 8(;08 8052, 44 7 8695 8739 878J 8826 8869 8913 8956 9000 9043 9087, 44 8 9131 9174 9218, 0261 93(15 9348 9392 9435I 9479 9522I 44 9| 9565 9609 9652 9696 9739 9783! 9826 9870i 99131 9957 43 N.j 1 1 1 2 ' 3 i 4 I 5 1 6 1 '7 1 8 1 9 ID. T A B L E OF LOOARITHMIC SINES, COSINES, TANGENTS, AND COTANGENTS, FOE EVEKY DEGREE AND MINUTE OF THE QUADEANT, 18 Sine. LOGARITHMIC SINES, COSINES, 0^ D. Cosine. TaiiK- I p. OotlLllg 4 5 7 8 9 10 11 12 13 14 15 16 17 IS 19 20 21 22 23 24 25 26 27 28 29 80 31 32 33 34 35 33 37 38 39 40 41 42 43 44 45 40 47 48 49 50 61 62 63 64 65 66 57 68 59 60 — 00 6.463726 .764756 .940.-47 7.0657b6 .162696 .241677 .30S824 .366816 .417968 .463725 7.505118 .642906 .677668 .609863 .639816 .667845 .694173 .718997 .742477 .764754 7.7 85943 .806146 ,825451 .843934 .861662 ,678695 .895085 .910879 .926119 .940842 7.965082 .968670 .9822-33 .995198 8.007787 .020021 .031919 .043501 .034781 .065776 8.076300 .066963 .097183 .107167 .116926 .126471 .135810 .114953 .1539J7 .162681 8.171280 .179713 .187985 .1SC102 .204070 .211895 .219581 .227134 .234567 .241865 5017.17 2934.85 2082.31 1615.17 1310.08 1115.78 966.53 852.54 762.03 683.83 629.81 579.37 536.41 493.38 437.14 4.38.81 413. ;2 391.35 371.27 35 J. 15 333.72 321.75 308.05 293.47 283.88 273.17 233.23 25i.99 243.38 237.33 229.80 222 73 216.03 203.81 203,90 198.31 193.02 188.01 183.23 178.72 174.41 170.31 166 39 162.65 153.08 155 i 152 33 141 21 143,22 143.33 140.54 137.83 135 -2.) 132 80 130.41 128.10 125.87 123.72 121.64 lo.ouoouo .000000 .000000 .000000 .000000 .000000 9.999999 .999999 .999999 .999999 9.999998 .999997 .999997 .999996 .999995 .999994 .999993 9.990992 .999991 .999990 .999988 .999987 .999985 .999983 9 999982 .999981 .999980 .999979 .999977 .999976 .999976 .999973 .999972 .999971 9.999969 .999968 .999961 .999959 .999956 .999964 9.999952 .999950 .999948 .999946 .999944 .999943 .999940 .999938 .999936 .999934 .00 .00 .00 .03 .09 .01 •.01 .01 .01 .01 .01 .01 .01 .01 .01 .01 .01 .01 .01 .01 .01 .01 .01 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .02 .Oi .03 .03 .03 .03 .03 .03 .03 .03 .03 .03 .03 .04 .04 .04 .04 — 00 6.463726 .764756 .940f:47 7.0657S6 .162696 .241878 .308833 .366817 .417970 .463727 7.505120 .642909 .677672 .609867 .639820 .667849 .694179 .719004 .743484 .764761 7 786951 .806165 .825460 .843944 .861674 .878708 .895099 .910894 .926134 .940858 7965100 .968869 .982263 .995219 8.007809 .020045 .031946 .043527 .034809 .065806 8.076531 •086997 .097217 .107202 .116963 .126510 .133851 .144990 .153952 .102727 8 171328 .179763 .188036 .196166 .204126 .211953 .219641 .227195 .234621 241921 5017.17 2934.85 2082.31 1615.17 1319.68 1115 78 996.53 852.54 762.63 683.88 023.81 579.38 536.42 499.33 467.15 438.82 413.73 331.36 371.28 331 17 333.73 321.76 303.08 2)3,43 283.90 273.18 2(33 25 254.01 215.40 237.35 223 81 222.75 216 10 209.83 203.92 198.33 193.05 188 03 183.27 178.74 174 44 170.34 166.42 112 68 159.10 135.08 152.41 149.27 146,27 143.3G 140.57 137.00 135 32 132 84 130.44 128.14 125 90 323.76 121.68 3.536374 .235244 .059153 2.934314 .837304 .758122 .691173 .633163 .582030 .536273 2.494880 .457091 .422328 .390143 .360160 .832151 .305821 .280997 .257616 .235239 2.214049 .193645 .174640 .156060 .138326 .121292 .104901 .069106 .073866 .059142 2.044800 .031111 .017747 .004761 1.992191 .979955 .968066 ,956473 .945191 .934194 1.923469 .913003 .902783 .693797 .883037 .873490 .864149 .853004 .846048 .837273 1.826673 .620237 .811964 .803844 .796674 .788047 .780359 .772805 .765379 ,758079 I Cosine. I D. | Sine. | D | Cotang. | D. | Tang. | M. 89° TANGENTS, AND COTANGENTS. 1° I'J M. I 1). (Josiiie. I). I Tang. I P. Cotang. b.-^41b56 1 .249033 2 .256094 3 .263042 4 .269S81 5 .276614 6 .283243 7 .289773 8 .296207 9 .302546 10 .308794 11 8.314904 12 .321027 13 .327016 14 .332924 15 .338753 16 .344504 17 .350181 18 .355783 19 .361315 20 .366777 21 8.372171 22 .377499 23 .382762 24 .387962 25 .393101 26 .398179 27 .403199 28 .408161 29 .413068 30 .417919 31 8.422717 32 .427462 33 .432156 34 .436800 35 .441394 36 .445941 37 .450440 38 .454893 39 .459301 40 .463665 41 8.467985 42 .472263 43 .476498 44 .480698 45 .484848 46 .488963 47 .493040 48 .497078 49 .501080 50 .505045 51 8.508974 52 .512867 53 .616726 54 .520551 55 .624343 56 .528102 57 .531828 58 .535523 59 .539186 60 .642819 119. G3 117. 6S 115.80 113.98 112.21 110.51) 108.83 107.21 105. (ia 104.13 102.66 J01.22 9l).82 98.47 97.14 95. 8 J 94.00 93.38 92.19 91.03 89.90 88.80 87.72 8G.G7 8-5.64 84.64 83 O'J 82. 7 L 8177 80.83 79.96 79.09 78.2.i 77.40 76 57 75.77 74.99 74 22 7.i.4i 72 73 72 0,1 71.2) 70.63 0J.9I 6J.21 68.59 07 94 67.31 66.0.) 60.08 05.48 64.89 64.31 63.75 63.19 62.64 62. n 6 1. .58 61.06 60.55 U.9UU934 .999932 .999929 .999927 .999925 .999922 .999920 .999918 .999915 .999913 .999910 9.999907 .999902 .999899 .999897 .999891 .999868 .999882 9.999879 .999676 .999873 .999870 .999867 .999664 .999861 .999858 .999854 .999851 9.999648 .999844 .999841 .999638 .999834 .999631 .999827 .999623 .999820 .990816 9.999812 .999609 .999805 .999801 .999797 .999793 .999790 .999786 .999782 .999778 9.999774 .999769 .999765 .999761 .999757 .999753 .999748 .999744 .999740 .999735 .04 .04 .04 .04 .04 .04 .04 .04 .04 .04 .04 .04 .04 .0. .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 05 .05 .05 .00 .06 .00 .00 .00 .00 .00 .00 .00 .00 .00 .06 .06 .03 .00 .07 .07 .07 .07 .07 .07 .07 .07 .('7 .07 .07 .07 .07 .07 .07 8.241921 .249102 .256165 .263115 .269956 .276691 .283323 .289856 .296292 .302634 8.315046 .321122 .327114 .333025 .338856 .344610 .350289 .355695 .361430 .366695 8.372292 .377622 .382669 366092 .893234 .396315 .403338 .406304 .413213 .416068 8.422669 .427618 432315 .436962 .441560 .446110 .450613 .455070 .459481 .463649 8.466172 .472454 .476693 ,460692 .465050 .489170 .493250 .497293 .501298 -50J267 8.509200 .513098 .516961 .520790 .524586 .528349 .532080 .535779 .539447 .543084 119.67 117.72 115.84 114.02 112.25 110.54 108.87 107.20 105.70 104.18 102.70 101.26 99.87 98.51 97.19 05.90 94.65 93.43 92.24 91.08 8J.95 88.85 87.77 80.72 85.70 84.70 83.71 82.76 81.82 80.91 80.02 79.14 78.30 77.45 76.63- 75.83 75.05 74.28 73.52 72.79 72.00 71.35 70.63 09.98 0.0.31 68 05 OS.Ol 67. .38 00.76 06.15 65.55 64.96 64.39 63 82 63 2 ! 6272 62.18 01.65 61.13 60.62 1.758079 60 .750698 59 .743835 .736885 .730044 .723309 .716677 .710144 .703708 .697366 .691110 1.084964 .678678 .672860 .666975 .661144 .666390 .649711 .644106 .636570 .633105 1.C27708 .622378 .617111 .611908 .606766 .601665 .596662 .691696 .666787 .661932 1.577131 .572382 .567665 .663038 .558440 .553890 .640367 .544930 .540519 .636151 1.531828 .627.546 .523307 .619108 .614950 .510630 .606750 .602707 .498702 .494733 1.490600 .486902 .483039 .479210 .475414 .471651 .467920 .464221 .460563 .456916 Cosine. | D | Sine. | D. | Cotang. 88^ D. I Tang. I M. 20 LOGARITHMIC SINES, COSINES, •2^ Sine. D. Cosine. D. Tang. I P. I Cotang. | u 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 60 61 62 63 54 55 66 67 68 59 [ 60 S.542S19 .646422 .549995 .663539 .657054 .660540 .663999 .667431 .570836 .574214 .577566 8.580893 .584193 .587469 .590721 .593948 .597152 .600332 .603489 .606623 .609734 8.612823 .615891 .618937 .621962 .624965 .627948 .630911 .633854 .636776 .639680 8.642563 .645428 .648274 .651102 .653911 .656702 .659475 .662330 .664968 .667689 8.670393 .673080 .676751 .678405 .681043 .683665 .686373 .691438 .693998 B 696543 .699073 .701589 .704090 .706577 .709049 .711507 .713952 .716383 .718800 60.04 59.55 59.06 58.58 58.11 57. B5 67.19 56.74 56.39 55 87 65.44 55.02 54.60 54.19 53.79 53.39 53.00 52.61 52.23 51.86 51.49 51.12 50.76 50.41 50.06 49.72 49.38 49 04 48.71 48.39 48.06 47.75 47.4:f 47.12 46.82 46.62 46.22 45.92 45.03 45.35 45.06 44 79 44.51 44 24 43.97 43.70 43.44 43.18 42.92 42.67 42.42 42 17 4192 41.68 41.44 41.21 40.97 40.74 40.51 40.29 9.999735 .999731 .999726 .999732 .999717 .999713 .999708 .999704 .999694 .999680 .999675 .999670 .999660 .999855 .999650 .999645 .999640 9.999635 .999629 .999624 .999619 .999614 .999603 .999697 .999592 .999566 9 999581 .999676 .999570 .999664 .999558 .999653 .999647 .999641 .999535 .999539 9.999534 .999518 .999512 Cosine. I D. .999500 .999493 .999487 .999481 .999475 .999469 9.999403 .999456 .999460 .999443 .999437 .999431 .999424 .999418 .999411 .999404 Sine. .07 .07 .07 .08 .08 .08 .08 .08 .08 .08 .08 .08 .08 .08 .08 .08 .08 .08 .08 .00 .09 .09 .00 .09 .09 .09 .09 .09 .09 .09 .09 .00 .09 .09 .09 .10 .10 .10 .10 .10 .10 .10 .10 .10 .10 .10 .10 .10 .10 .10 .10 .11 .11 .11 .11 .11 .11 .11 .11 .11 87° 8.543084 .546691 .650268 .553617 .557336 .660828 .564291 .567727 .671137 .574620 .577S77 8.681208 .584514 .587795 .691061 .594283 .597493 .600677 .603839 .606978 .010094 8.6131S9 .616362 .619313 .622343 .625352 .628340 .631308 .634366 .637164 .640093 8 642982 .646to3 .646704 .651537 .664.363 ■657149 .659938 .662689 .665433 .668160 8.G70870 ■673563 .676239 .676900 .661644 .084172 .686784 .689361 .691963 .694539 8 697081 .699617 .702139 .704646 .707140 .700618 .712083 .714534 .716972 719396 Cotang. 60.12 59.62 59.14 58.66 58.19 87.73 57.27 56.82 56.38 55.95 65.52 55.10 54.68 54.27 53.87 53.47 53.08 52.70 52.32 51.94 51.58 51.21 50.85 50.50 50.15 49.81 49.47 49.13 48.80 48.48 48.16 47.84 47.53 47.22 46.91 46.61 46.31 46.02 45.73 46.44 45.10 44.88 44.01 44.34 44.07 43.80 43.64 43.28 48.03 42.77 42.62 42 28 42,03 41.79 41.55 41.32 41.08 40.85 40.62 40.40 1.456916 .463309 . .449732 .446183 .442664 .439172 .435709 .432273 .428603 .425480 ' .423123 1.418792 .415486 .413205 .408949 .405717 .402608 .399323 .396161 .393022 .389906 1,386811 .3.S3738 .380687 .377667 .374648 .371660 .368692 .365744 .362816 .369907 1.357018 .354147 .851296 .348463 .345648 .343851 .340072 .337311 .334567 .331840 1 320130 .32 i437 .323761 .321100 .318456 .815628 .313316 .310619 .306037 .305471 1.302919 .300383 .297861 .295364 .293860 .290383 .287917 .285465 .283028 .280604 D. I Tang. TANGENTS, AND COTANGENTS. 3^ 21 U. I Sine. I 1). I Cosine. | D. | 'I'ang. | D. 1 Cotang. I S.TlbbOO .721204 .723595 .725972 .726337 .730688 .733027 .736354 .737667 .742259 8.744536 .746802 .749055 .751297 .753528 .765747 .757955 .760151 .762337 .764511 8.760675 .766828 .770970 .773101 .775223 .777333 .779434 .783624 .783005 .785675 8.787736 .789787 .791828 .793859 .795881 .797894 .799897 .801692 .803876 .805852 8.807819 .609777 .811726 .813667 .815599 .817522 .819436 .821343 .823240 .825130 8.827011 .828884 .830749 .832607 .834456 .836297 .838130 .839966 .841774 .843585 40.06 39.84 39.62 39.41 39.19 38.98 38.77 38.57 33,36 3S.1S 37.98 37.76 37.56 37.37 37.17 36.98 36.79 36.61 36.42 36 24 36.03 35. 8S 35.70 35.53 35.35 35.18 35.01 34.84 34.67 34.51 .34.34 3*. 18 34.02 33.88 33.70 33.54 33.39 33.23 33.08 32.93 32.78 32.63 32.49 32.34 32.19 32.05 31.91 31.77 3163 31.49 31.-35 31.22 31.08 30.95 30.82 30.69 30.56 30.43 30.30 30.17 0.999404 .990398 .999391 .999384 .999378 .999371 .999364 .999357 .999350 .990343 .999336 9.999329 .999322 .999315 .999308 .999301 .999294 .999286 .999279 .999272 .999265 9.999257 .999250 .999242 .999235 .999227 .999220 .999212 .999205 .999197 .999189 9.999181 .999174 .999166 .999158 .999150 .999142 .999134 .999126 .999118 .999110 9.999102 .999094 .999086- .999077 .999069 .999061 .999053 .999044 .999036 .999027 9.999019 .999010 .998993 .998984 .998976 .998967 .996958 .998960 .996941 .11 .11 .11 .11 .11 .11 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .13 .14 .14 .14 .14 .14 .14 .14 .14 .14 .14 .14 .U .14 .14 .14 .15 .15 .15 8.719396 .721806 .724204 .726588 .728959 .731317 .733663 .735996 .738317 .740626 .742922 8.745207 .747479 .749740 .761969 .754227 .756453 .758668 .760872 .763065 .765246 8.7S7417 .769678 .771727 .773866 .775995 .776114 .780222 .782320 .784408 .786486 8.788554 .790613 792662 .794701 .796731 .798762 .800763 .802765 .804758 .806742 8.608717 .810683 .812641 .814569 .816529 .818461 .820364 .822298 .824205 .826103 8.827992 .829874 .831748 .833613 .835471 .837321 .839163 .840998 .842825 .844644 40.17 39.95 39.74 39.52 39.30 39.09 38 8,J 38.68 38.48 38.27 38.07 37.87 37.68 37.49 37.29 37.10 36.92 38.73 38.55 36.33 38.18 38 00 35.83 35.65 35.48 35 31 35. U 34 97 34.30 34.64 34.47 31.31 34.14 33.99 33 83 3.5.63 33.52 3)3 7 33 22 33.07 32.92 •32.77 32.62 32 48 32 33 32 19 32.05 31.91 31.77 31.63 31.59 313) 31.23 31.10 30.96 30.83 30.70 30 57 30.45 30.32 1.280004 .276194 .275796 .273412 .271041 .268663 .266337 .264004 .261683 .259374 .257078 1.284793 .252621 .250260 .246011 .246773 .243547 .241332 .239128 .236935 .234754 1.232563 .230422 .228273 .226134 .224005 .221686 .219778 .217680 .216592 .213614 1.211446 .209367 .207338 .205299 .203209 .201248 .199237 .197235 .195242 .193258 1.191283 .189317 .187359 .185411 .163471 .181539 .179616 .177702 .175795 .173897 1.172008 .170126 .168252 .166387 .164629 .162679 .160837 .159002 .107175 .155366 Cosine. D. Sine^l p. I Cotang. | D. | Tang. | M 86° 22 "li'l Sine. I LOGARITHMIC SINES, COSINES, 40 D I Oo.siiie. I D. I Tang. | I). | Cotang. | C 7 8 9 10 11 12 13 14 15 16 1- 13 19 20 21 22 23 24 25 28 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 «0 51 52 53 54 55 06 69 00 S.b435b6 .845387 .847183 .848971 .850751 .852525 .854291 .856049 .857801 .859546 .861283 8.863014 .864738 .866455 .868165 .809868 .871565 .873255 .874938 .376615 .878285 8.879949 .881607 .883258 .884903 .886542 .888174 .889801 .801421 .893035 .894643 8.896246 .897842 .899432 .901017 .902596 .904169 .905736 .907297 .908853 .910404 8 911949 .013488 .915032 .916550 .918073 .919591 .921103 .922310 .924112 .925009 8.927100 .928587 .930068 .931544 .933015 .931481 .935942 .937398 .938850 .940290 30.05 2J.92 ■29.80 '29.07 29.55 2J.43 2.J3L 23.19 2i).07 28.96 28.84 28 73 23. GL 28, .50 28,3.9 2S.23 28,17 23.(13 27.95 27.83 2r.73 27.63 27.52 27.42 27.31 27.21 27.11 27.00 2). 90 23.80 26.70 21.60 23.51 23.41 23 31 2,i.22 2i.l2 21.03 2") g-i 25 84 2'),75 25.03 25.60 25.47 2).. 38 25 29 25.20 25.12 25.03 24.94 24.83 24.77 24,69 24.6(1 ■24 52 24.43 24.35 24 27 24.19 24.11 9.998941 .998932 .998923 .998914 .998905 .998887 .998878 .998861 9.998841 .998832 .998823 .908813 .998804 .998795 .998785 .998770 .908706 998757 0.998747 .998738 .998728 .098718 .908708 .998099 .99,S689 .998679 .908669 .998069 9,908649 .998630 .998629 .998010 .998609 .998599 .998689 .998578 .998568 .098658 .998537 .998527 .998510 .998506 .998495 .998485 .908474 .998404 998453 9.998442 .998431 .998421 .998410 .998399 .998388 .998.377 .998306 .998355 .098344 .15 .15 .15 .15 .15 .15 .15 .15 .15 .15 .15 .15 .15 .10 .10 .16 .16 .16 .16 .16 .16 .16 .16 .16 .10 .16 .16 .10 .16 .17 .17 .17 .17 .17 .17 .17 .17 .17 .17 .1! .17 .17 .17 .17 .18 .U .18 .18 .18 .18 .18 .18 .18 .18 .18 .18 .18 .18 .18 .18 8.844044 .846455 .848200 .860057 .851846 .853628 .855403 .857171 .858932 .860666 .862433 8.S64173 .865906 .867632 .869351 .871004 .872770 .874409 .876162 .677849 .879529 8.881202 .882869 .884530 .886165 .667833 .889476 .891112 .802742 .804366 .895084 3 897596 .899203 .900803 .902108 .9039S7 .905570 .907147 .908719 .910285 .911846 3.013401 .914951 .910495 .018034 .919508 .921096 .922019 .924130 .925040 ,927156 B 928058 .930155 .931647 .933134 .934610 .936003 .037565 .930032 .940494 .041952 30.19 30.07 29.95 29.82 29.70 29.58 29.46 29.35 29.23 29.11 29.00 28.88 28.77 28.66 28 54 28.43 28.. 32 28.21 28.11 28.00 27.89 27.79 27.68 27.58 27.47 27.37 27.27 27.17 27.07 26.97 28.87 26.77 26.67 26.58 26.48 20.38 26 2n 20.20 20.10 26.01 25 92 25.83 25.74 25.65 25.56 25.47 25.88 25.. 30 25.21 25.12 25.03 24.95 24.86 24.78 24.70 24.61 24.53 24.45 24.37 24.29 1.155356 .153545 .151740 .149943 .148154 .146372 .144597 .142E29 .141003 .139314 .137507 1.135E27 .134094 .132008 .130049 .12f:936 .127230 .125531 .123838 .122151 .120471 1 116798 .117131 .115470 .113815 .112167 .110524 .10£f68 .107258 .105C34 .104016 1.102404 .100797 .009197 .097002 .090013 .094430 .092853 .091381 .0(0715 .018154 1 Of 0599 .085049 .083505 .OEioeo .080432 .07f:C04 .077381 .075f:64 .074351 o;aC44 1.071342 .069845 .068353 .066860 .006384 ■ .063907 .062435 .060908 .069506 .058048 I Cosine. D. I Sine. I D. I Cotiino;. | D. I Tang. |1\1. TANGENTS, AND COTANGENTS. 2■^ M. Sine. D. I Cosine. U- I Tang D. I Cotang 9 10 n 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 SO 31 32 33 34 35 36 37 33 39 40 41 42 43 44 45 46 47 48 49 SO 53 64 55 66 57 58 59 60 8.9-111296 .041738 .943174 .944606 .946034 .947456 .!'4S574 .950287 .951096 .953100 .954499 8.955894 .9572S4 .95E6T0 .960052 .961429 .962E01 .964170 .905534 .906f:93 -9GE249 8.969600 .970947 .972:e9 .973028 .974962 .970293 .977019 .978941 .9E0259 .981573 8.S82[83 .984169 .9!:5491 .980789 .SE80S3 .989374 .C90C00 .991943 .9!. 3222 .994497 8.;i'.'5rC8 .997036 .698299 .999560 9.000810 .002069 .003318 .004563 .005105 .007044 b.008278 .009510 .010737 .011952 .013182 .014490 .015013 .01G!:24 .018031 .010235 24.03 23 94 23,87 23.79 23.71 23.63 23.5.5 23.48 23.40 23.32 23,25 23.17 23.10 23,02 22,95 22 88 22,80 22,73 22 06 22,59 22,52 22 44 22,38 22 31 22 24 22.17 22 10 22 03 21 97 21.90 21.83 21.77 21,70 21,63 21,57 21.50 21.44 2138 21.31 21,25 21 19 21.12 21.00 21.00 20 94 20 87 20,82 20,76 20,70 2(1 64 20,58 20.52 211.46 20,40 20,:i4 20,2.1 20,2 i 20.17 20.12 20,06 9.998344 .998333 .998322 .998311 .998300 .998269 .998277 .998266 .998255 .998243 .998232 9.996220 .998209 .998197 .9981f6 .998174 .998163 .998151 .998139 .998128 .1,98116 9.998104 .998092 .998060 .996008 .996056 .996044 .998032 .998020 .996008 .997996 9.997985 .997972 .997959 .997947 .997935 .997922 .997910 .997697 .997685 .997672 9.997860 .997647 .997635 .997822 .997609 .997797 .997764 .997771 .997758 .997745 9.997732 .997719 .997706 .997093 .997080 .997067 .997654 .997041 .997028 .'97014 8.941952 .943404 .944852 .940295 .947734 .949168 .960597 .962021 .963441 .954656 .956207 8,957074 .959075 .900473 .961660 .903255 .964039 .900019 .967094 .968706 .970133 8.971490 .972:: 55 .974209 .975500 .970906 .978243 .979560 .980921 .982251 .983577 8.984899 .966217 987532 .988842 .990149 .991451 .992750 .994045 .995337 .990024 8.997908 .999168 9.000465 .001738 .003007 .004272 .005534 .006792 .006047 .009298 9.010546 .011790 .013031 .014268 .015502 .016732 .017959 .019113 .020403 .021020 24,21 24.13 24.05 23.97 23.90 23,82 23,74 23,66 23.58 23,51 23,44 23.37 23,29 23,22 23,14 23,07 23.00 22,93 22,80 22 79 22 71 22,65 22,57 22.5: 22,44 22,37 22,30 22 23 22,17 22 10 22,04 21,97 21,91 21,84 21.78 21,71 21.65 21.68 21.52 21.46 21.40 21,34 21,27 21,21 21.15 21,09 21.03 20,97 20.91 20.85 20,80 2I).74 20.68 20,62 20.66 20.51 20,45 20,39 20,33 20.28 1.058U48 .050596 .065148 .053705 ,052266 .050632 .049403 .047979 .046559 .045144 .0437S3 1.042.326 .040925 .039527 .038134 .030745 .005301 .030961 .032000 .031234 .029667 1.028504 .027145 .025791 .024440 .023094 .021752 .020414 .019079 .017749 .016123 1,015101 .010783 .012408 .011158 .009651 .006549 .0072C0 .005955 .004003 .000376 1.002092 .000612 0.099535 .998262 -.996998 .995728 .994466 .990208 .991953 .990702 0.969454 .986210 .966969 .965732 .984498 .963268 .982041 .960817 ,9795',.7 ,9710(0 I Cosine D, I Sine. I 1). I Cotang 84= D. I Ti I M. 24 LOGARITHMIC SINES, COSINES, 6° Sine^ I D 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 36 37 38 39 iO, 41 43 43 44 45 46 47 48 49 50 51 62 63 64 55 66 67 58 59 60 9.01»235 .020435 .021632 .022825 .024016 .025203 .026386 .027567 .028744 .029918 -031089 9.032267 .033421 .034682 .035741 .036896 .038048 .039197 .040342 .041486 .042625 9.043762 .044895 .046026 .047154 .048279 .049400 .050519 .051635 .052749 .053859 9.054966' .056071 .057172 .058271 .059367 .060460 .061661 .062639 .063724 .064606 9.065885 .066962 .068036 .069107 .070176 .071242 .072308 .073366 .074424 .07.5480 9.076533 .077583 .078631 .079676 .080719 .081759 .082797 .0.S3632 .084664 .085894 20.00 10.95 19.89 19.84 19.78 19.73 19.67 19.62 19.57 19.51 19 47 19.41 19..36 19.30 19.25 19.20 19.15 19.10 19.05 18.99 18.94 18. 8J 18.84 18.79 18.75 18.70 18.65 18.00 18.55 18.50 18.45 18.41 18.36 18.31 18.27 18.22 18.17 18.13 18.08 18.U4 17.99 17.94 17.90 17.83 17.81 17.77 17.72 17.68 17.03 17.59 17.55 17.50 17. 40 17.42 17.38 17 33 17.29 17.25 17.21 17.17 Cosine. 9.997614 .997601 .997588 .997674 .997561 :997647 .997534 .997520 .997507 .997493 .997480 9.997466 .997452 .997439 .997425 .997411 .997397 .997383 .997369 .997355 .997341 9.997327 .997313 .997299 .997285 .997271 .997267 .997242 .997228 .997214 .997199 9.997165 .997170 .997156 .997141 .997127 .997112 .997098 .997083 .097068 .997053 9.997039 .997024 .997009 .996994 .996979 .996964 .996949 .996934 .996910 .996904 9.996689 .990874 .996658 .996643 .996828 .996812 .996797 .996782 .996706 .996751 D. Tans. .22 .22 .22 .22 .22 .22 .23 ,23 .2:3 .23 .23 .23 .23 .23 .23 .23 .23 .2! .23 .23 .23 .24 .24 .24 .24 .24 .24 .24 .24 .24 .24 .24 .24 .24 .24 .24 .24 .24 .25 .25 .25 .2) .25 .2) .25 9.U21620 .022834 .024044 .025251 .026465 .027655 .026852 .030046 .031237 .032425 .083609 9.034791 .036969 .037144 .038316 .039485 .040651 .041813 .042973 .044130 .045284 9.046434 .047582 .048727 .049869 .051008 .052144 .053277 .054407 .055535 .056659 ) 057781 .068900 .060016 .061130 .062240 .063348 .064453 .065556 .066655 .067752 D.068846 .069938 .071027 .072113 .073197 .074278 .075356 .076432 .077505 .078576 } 079644 .080710 .081773 .082633 .083691 .084947 .066000 .087050 .088098 .089144 D. I Cotang. | 20.23 20.17 20.11 20.06 20.00 19.95 19.90 19.85 19.79 19.74 19.09 19.64 19.58 19.53 19.48 19.43 19.38 19.33 19.28 19.23 19.18 19.13 19.08 19.03 18.98 18.U3 18 83 18.81 18.79 18 74 18.70 18 63 18.60 18.55 18 51 18 43 18 42 18 37 18.33 18.28 18 24 18.19 18.15 18.10 18.08 18.02 17.07 17.93 17.83 17.84 17.80 17.76 17.72 17.07 17.63 17.50 17.55 17.51 17.47 17.43 0.978360 6( .977166 69 .975956 68 .974749 57 .973545 66 .972345 55 .971148 54 .969954 53 .968763 62 .967576 61 .966391 60 D.966209 49 .964031 48 .962866 47 .961684 46 .960515 45 .959349 44 .958187 43 .967027 42 .966670 41 .954716 40 0.953666 39 .952418 38 .951273 37 .950131 36 .948992 35 .947856 34 .9467^3 33 .945593 32 .944465 31 .943341 30 0.942219 29 .941100 28 .939984 27 .936870 26 .937760 25 .936662 24 .936647 23 .934444 22 .933345 21 .932248 20 931154 19 .930002 18 .938973 17 .927887 16 .926603 15 .925723 14 .924644 13 .923568 12 .922495 11 .921424 10 0.920356 9 .919290 8 .916227 7 .917167 6 .916109 6 .915053 4 .914000 3 .912950 2 .911903 1 .910656 Tang. 1 M. I Cosine. | D. | Sine. | D- | Cotang. | D. 83° TANCJEiNTri, AND COTANGENTS. 7-= 51 02 53 Di- ss 66 57 68 59 60 Sine. D. 9.085804 .086922 .087947 .088970 .089990 .091008 .092024 .093037 ,094047 .095056 .09G062 9.097065 .098066 .099065 .100062 .101056 .102048 .103037 .104025 .106010 .105992 9.106973 .107951 .106927 .109901 .110873 .111842 .112809 .113774 .114737 .115698 9.116656 .117613 .118567 .119519 .120469 .121417 .122.362 .123306 .124248 .125187 9.120125 .127000 .127993. .12692.5 .129854 .130781 .131706 .132630 .133551 .134470 9.135367 .136003 .137216 .138128 .139037 .139944 .140C50 .141754 .142655 .143555 I Ciiajiic. I 1). I T;uig. I D. I Cutaiis. | 17.13 17.09 17.04 17.00 IC.'JG 10.92 10.88 16.84 16.80 16.76 16,72 1U.6S li).65 l.i.Ol Id.. 57 IJ o-i 16.49 i:;.45 10.41 10.38 10.34 16-30 10 27 lO.-ii ]0,l» 10.16 15 12 18.08 Li U3 10.01 15.97 . 15.94 15.90 15,87 15 8-i 15.8J 15.76 15,73 15.03 15.06 15,62 1.5.59 15.56 1.5.52 15.49 1.5.45 15.42 15.39 15.35 15..32 15.29 1.5.25 15.22 1-5.19 15.16 15.12 15.09 15 00 15.03 15.00 9,996751 .996735 .996720 .996704 .996688 .996673 .996657 .996641 .990625 .996010 996.594 9,996:78 .990562 .996546 .996530 .996614 .996498 .996482 .996465 .996449 .990433 9,996417 .990400 ,996384 .996368 .996351 ,996335 .996318 .990302 .996285 .996269 9.996252 ,996235 .996219 .996202 .996185 .996168 .996161 .996134 .996117 .996100 9.996063 .996066 .996049 .996032 .990015 .995998 .995980 .995963 .996946 .99.3928 9.995911 .996694 .995876 .995859 .995841 .995823 .996606 .995788 .995771 .995753 ,27 ,27 ,27 ,27 .27 ,27 ,27 ,27 ,27 .27 .27 .27 .27 •2; ,27 ,27 ,27 ,28 .28 .28 .28 .28 .28 .2.S ,2S ,28 .28 .28 .28 .28 .29 .2a .29 .29 .29 ,2J ,21 ,2,) ,2 1 2) ,21 9.089144 .090167 .091228 .092266 .093302 ,094336 .096367 .096396 .097422 .096446 .099468 9,100487 .101504 .102619 .103532 .104542 .105650 .106566 .107559 .108660 .109569 9.110656 .111551 .112643 ,113633 .114521 .116507 .116491 .117472 .116452 .119429 9,120404 .121377 122348 .123317 .124264 .125249 .126211 .127172 .128130 .129087 9,130041 .130894 .131944 .132693 .133639 .134764 .135726 .136607 .137006 .138542 9.139476 .140409 .141340 .142269 .143196 .144121 .146044 .145966 .146865 .147603 17.38 17..S4 17.30 17,27 17.22 17.19 17.15 17.11 17.07 17 03 10.99 10,95 16.91 10.87 10.84 10.80 10.76 10.72 10.09 10.63 10.61 36,58 16.54 10.50 10.40 16.43 16.39 16.36 10,32 16.29 10.25 16,22 16.18 10.13 16.11 16.07 16.04 16.01 13.97 15.94 15.91 15.87 15.84 15,81 15.77 15 74 15,71 15,67 35,64 35 61 15, .58 15,53 15,51 15,48 35,45 35,42 15,39 1.5.35 15..S2 15.29 0,910866 I 60 .909813 59 .908772 I 68 .907734 .905664 .904633 .903605 .902578 .901564 .000532 0.S995I3 .898496 .897481 .896468 .695468 .894450 .893444 .692441 .891440 .890441 0.889444 .886449 .687457 .886467 .865479 .684493 .883509 .882528 .681548 .680571 0.879596 .876623 .877662 .876683 .875716 .674761 .873769 .872628 .671870 .670913 0.869969 .669006 .868056 .667107 .666161 .665216 .664274 .8633.33 .862395 .861468 0.860524 .669691 .656660 .867731 .856804 ,865879 .864966 .854034 .663115 .852197 49 48 47 46 45 44 43 42 41 40 39 38 37 30 36 34 I 31 30 29 28 27 26 26 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 Cosine, I D. | Sine. | D. | Cotang. | D. | Tnii.g. | M 3* 82^ LOGAEITHMIC SINES, COSINES, Sine. I D I Onship- I D. I Tang. | i). | Cotans U 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 40 41 43 43 44 45 48 47 48 49 50 51 52 53 64 55 56 58 69 60 9.143565 .144453 .145349 .146243 .147136 .148026 .148915 .149802 .160686 .161569 .152451 9.153330 .154208 .166083 .155957- .156830 .157700 .158569 .159436 .160301 .161164 9.162025 .162885 .163743 .164600 .165454 .166307 .167159 .168008 .168856 .169702 9.170547 .171389 .172230 .173070 .173908 .174744 .175578 .176411 .177242 .17S072 9.178900 .179726 .180551 .181374 .182196 .183018 .183834 .184651 .185466 .18G2tO 9.187092 .187903 .1SSV12 .189619 .190325 .191130 .191933 .192734 .193534 .194333 14.96 14.93 14.90 1J.87 14.84 14.81 14.78 14.75 14.72 14.Uil 14.66 14.03 14.GJ 14.57 14.54 14.51 14 48 14.45 14.42 14.39 14.3S 14.33 14.30 14.27 14.24 14.22 14 19 14.16 14.13 14.10 14.07 14.05 14.02 13.99 13.96 13.94 13.01 13.88 13.86 13.83 13.80 13.77 13.74 13.72 13.69 13.66 13.64 13.61 J3..W 13.50 13.53 13,51 13.48 13.46 13.43 13 41 13.38 13.36 13.33 13.30 9,995753 .995735 .995717 .995699 .995081 .995664 .995646 .995628 .995610 .996591 .996673 9.995555 .995537 .996519 .995501 .996482 .995464 .995446 .995427 .995409 .995390 9.995372 .995353 .995334 .995316 .995297 .995278 .996260 .995241 .995223 .995303 9.995184 .995165 .995146 .996127 .995108 .996069 .995070 .995051 .996032 .995013 9.994D93 .994974 .994955 .994935 .994916 .994877 .994867 .994838 .994818 ).994798 .994779 .994769 .994739 .994719 .994700 .9946S0 .994000 .094040 .994620 .30 .30 .30 .30 .30 .30 .30 .30 .30 .CO .30 .30 .30 .20 .31 .31 .31 .31 .31 .31 .31 .31 .31 .31 .31 .31 .31 .31 .32 .32 .32 ..32 .32 .32 .32 .32 .32 .32 .32 .32 .32 .32 .32 .32 .32 .33 .33 .33 .33 .33 .33 .33 ..33 .33 .33 .33 .33 .33 .33 .33 9.147803 .148718 .149632 .150644 .151454 • .162363 .153269 ■.164174 .156077 .155978 .150877 9.157775 .158671 .159665 .160457 .161347 .162236 .163123 .164008 .164892 .165774 9.166654 .167532 .168409 .169284 .170157 .171029 .171899 .172767 .173034 .174499 9 176302 .176224 .177084 .177942 .178799 .179655 .180508 .181300 .182211 .183059 9.1S3907 .184752 .185597 .180439 .187280 .188120 .188958 .189794 .190629 .191462 9 192294 .193124 .193953 .194780 .196606 .196430 .197253 .198074 .198S94 .199713 15.26 15.23 15.20 16.17 15.14 15.11 15.08 15.05 15,02 14,99 14,96 14,93 14 90 14,87 14,84 14,81 14.79 14.70 14.73 14.70 14.07 14.64 1401 14.58 14,55 14,53 14,50 14.47 14.44 14.42 14.39 14.36 14.33 14.31 14 28 14.25 14.23 14.20 14.17 14.15 14.12 14,09 14 07 14 04 14,02 13,99 13,90 13,93 13,91 13.89 13.86 13.84 13 81 13.79 13.76 13.74 13.71 13.69 13.66 13.64 0.852)97 60 .851282 59 .850368 58 .849456 67 .848546 56 .847637 55 .846731 64 .845826 63 .844923 52 .844022 51 .843123 50 0.842225 49 .841329 48 .640435 47 .839543 46 .838653 45 .837764 44 .836877 43 .E33992 42 .836108 41 .834226 40 0,833346' 39 .832468 38 .831591 37 ,630716 36 .829843 35 .828971 34 .828101 33 .827233 32 .820366 31 .825501 30 0.824038 29 .823770 28 .822916 27 .822058 20 .821201 25 .820346 , 24 .819492 23 .818640 22 .817789 21 .£16941 20 0,816093 19 .815248 18 .814403 17 .f.13501 16 .812720 16 .SllStO 14 .811042 13 .cio:o6 12 .809371 11 .808538 10 0.807703 9 .800876 8 .806047 7 .805220 ■ 6 .804394 6 .803570 4 .802747 3 .801920 2 .801100 1 ,800287 Co^^ine. n. Sine. I U. I Cotan;;, | D. | Tang. TANGENTS, AND COTANGENTS. 9° 27 M. Sine. I P. I Cosine. | ]). | u 'J.1J4332 1 .195139 2 .195n ifiTTRT^T)^ «*taS!: