' -M f^I ' 1 . ( KEY TO A TREATISE ON SURVEYING BY / SAMUEL ALSOR AUTHOR OP "A TREATISE ON ALGEBRA," ETC. PHILADELPHIA: E. C. & J. BIDDLE, No. 508 MINOR STREET, (Between Market and Chestnut Sts., west of Fifth St.) 1857. Entered according to Act of Cot,grcsH, in the year 1857, by E. C. & J. BIDDLE, in the Clerk's OfBce of the District Court of the United States for the Eastern District of Pennsylvania. STEREOTYPED BY L. JOHNSON & CO. PHILADELPHIA. LC Control Nuinber tmp96 025490 KEY TO ALSOP^S SURVEYING. CHAPTEE III. PLANE TRIGONOMETRY. Article 137. Ex. 3. Here, (Eule 3,) As AB 59.47 A. C. 8.225702 : BC 48.52 1.685921 ; ; rad. 10.000000 : tan. A 39° 12' 36^' 9.911623 And (Rule 2) ' As COS. A 39° 12' 36''. A. C. 0.110791 : rad. 10.000000 :: AB 1.774298 : AC 76.75 1.885089 Ex. 4. Here, (Rule 1,) As AC 97.23 A.C. 8.012200 : BC 75.87 1.880070 : : rad. 10.000000 : sin. A 51° 17' 22" 9.892270 3 p. 66.] KEY TO ALSOP'S SURVEYING. [Chap. III. And (Rule 2) As rad. A. C. 0.000000 : COS. A 51° 17' 22" 9.796148 :: AC 1.987800 : i\B 60.81 1.783948 Ex. 5. Here, (Rule 3,) As tan. A 42° 19' 24" A. C. 0.040636 : rad. 10.000000 : : BC 25.54 1.407221 : AB 28.045 1.447857 And (Rule 1) As sin. A A. C. 0.171783 : rad. 10.000000 :: BC 1.407221 : AC 37.932 1.579004 Ex. 6. Here A = 90° - C = 17° 17' 51" Rulel. As rad. A. C. 0.000000 : sin. A 17° 17' 51" 9.473243 :: AC 495 2.694605 : BC 147.18 2.167848 And (Rule 2) As rad. A.C. 0.000000 : COS. A 9.979900 :: AC 2.694605 : AB 472.612 2.674505 Ex. 7. Rule 3. " As rad. A. C. 0.000000 : tan. A 42° 8' 45" 9.956659 :: AB 63.2 1.800717 : BC 4 57.20 1.757376 Chap. III.] PLANE TRIGONOMETRY. [Pp. 66-69. And (Rule 2) As COS. A : rad. :: AB : AC 85.24 A. C. 0.129925 10.000000 1.800717 1.930642 Article 138. Ex. 2. AC = x/AB^ + AC^ = x/36 + 64 = ^/100 = 10. Ex. 3. BC = v/(AC + AB) . (AC - AB) = ^/77.37 x 18.47 = x/1629.0239 = 37.8. Ex. 4. 49.27 + 37.42 = 86.69 49.27 - 37.42 = 11.85 Ans. 32.05 log. 1.937969 " 1.073718 2 )3.011687 1.505843 Article 139. Ex. 3. Here, (Rule 1,) AsAB : AC : : sin. C : sin. B A And (Rule 2) - As sin. C : sin. A :: AB : BC 32.774 37.25 42.59 57° 29' 15'' 74° 36' 53" 47° 53' 52" A. C. 8.428874 1.629308 9.925969 9.984151 A.C. 0.074031 9.870375 1 .571126 1.515532 Ex. 4. Here C = 180 - A + B == 125° 57' 14" And (Rule 2) As sin. C : sin. B :: AB : AC 125° 57' 14" 24° 15' 17" 325 164.93 A. C. 0.091788 9.613624 2.511883 2.217295 p. 69.] KEY TO ALSOP'S SURVEYING. [Chap. III. Eule 2. As sin. C : sin. A :: J\B 29° 47' 29" A.C, . 0.091788 9.696219 2.511883 : BC 199.48 2.299890 Ex. 5. As AC : .AB : : sin. B 106.49 127.54 52° 27' 18'' A.C. 7.972691 2.105646 9.899205 : sin. C 108° 16' 3" 9.977542 As sin. B : sin. A :: AC 19° 16' 39" A.C. 0.100795 9.518703 2.027309 : BC 44.34 2.646807 Ex. 6. Eule 1. As AC : AB : : sin. B 398.47 ' 527.63 43° 29' 11" A.C. . 7.399605 2.722330 9.837703 : sin. C or 65° 40' 44" 114° 19' 16" 9.959638 Eule 2. As sin. B : sin. A :: AC C acute. 43° 29' 11" 70° 50' 5" A.C. 0.162297 9.975237 2.600395 : BC 546.93 C obtuse. , 2.737929 As sin. B : sin. A : : AC 22° 11' 33" A.C. 0.162297 9.577169 2.600395 : BC G 218.71 2.339861 Chap. III.] PLANE TRIGONOMETRY. [P. 71. " Article 140. Ex. 1. As AB + BC 1875.10 A.C. 6.726976 : AB-BC 176.04 2.245611 C + A : : tan. — - — 2 57° 8' 20'' 8° 16' 10" 10.189790 C-A : tan. 2 9.162377 C = 65'' 24' 30" A = 4«° 52' 10" Hule 2, Art. 139. As sin. A A. C. 0.123083 : sin. B 65° 43' 20" 9.959787 :: BC 849.53 2.929179 : AC 1028.13 3.012049 Ex. 3. As AB + AC 373.39 A. C. 7.427838 : AB~AC 61.47 1.788663 C + B : : tan. — - — A G^"" 20' 30" 19° 43' 42" 10.338123 C-B : tan. — A 9.554624 C = 85° 4' 12" B = 45° 36' 48" As sin. C 85° 4' 12" A. C. 0.001610 : sin. A 49° 19' 9.879855 : : AB 217.43 2.337320 : BC 165.495 2.218785 p. 73.] KEY TO ALSOP'S SURVEYING. [Chap. Ill Article 141. Ex. 2. As AC A. C. 6.270585 : BC 3.964217 : : rad. 10.000000 ; tan. X 59° 47' 5" 45° 10.234802 tan. X — 45° 14° 47' 5'' 9.421483 A + B tan. 2 58° 21' 18" 23° 10' 58" 10.210207 A-B tan. ^ 9.631690 A 81° 32' 16" As sin. A A. C. 0.004754 : sin C. 63° 17' 24" 9.950993 :: BC 3.964217 : i\B 8317 3.919964 Ex. 3. As AB A. C. 8.036575 : BC 2.416347 : : rad. 10.000000 : tan. x 70° 35' 9" 45° 10.452922 tan. X — 45° 25° 35' 9" 9.680169 A+ C tan. 2 25° 9' 12° 40' 11" 9.671634 A- C tan. — - — 2 9.351803 A 37° 49' 11" As sin. A A. C. 0.212413 : sin. B 129° 42' 9.886152 :: BC 2.416341 : AC 327.27 2.514912 8 ruAP. III.] PLANE TRIGOXOME] Article 142. CRY. [: Ex. 2. Here As AB 214 A.C. 7.669586 : AC + BC 362 2.558709 : : AC - BC 28 1.447158 : ^T)-DB 47.364 1.675453 Whence AD = 1 L30.682 and BD = 83.318. Then (Art. 137, Rule 2,) As AC 195 A.C. 7.709965 : AD 130.682 2.116216 : : rad. 10.000000 : COS. A 47° 55' 13'^ 9.826181 And AsBC 167 A. C. 7.777284 : BD 83.318 1.920739 : : rad. 10.000000 : CQS. B 60° 4' W 9.698023 Ex. 3. As AB 51.67 A. C. 8.286762 : AC + CB 71.11 1.851931 : : AC - CB 16.79 1.225051 : AD-DB 23.107 1.363744 Hence AD = 37.3885 and BD = 14.2815. Then As AC 43.95 A. C. 8.357041 : AD 37.3885 1.572738 : : rad. 10.000000 : cos. A 31° 42' 42'' 9.929779 And AsBC 27.16 A. C. 8.566070 : BD 14.2815 1.154774 : : rad. 10.000000 : cos. B 58° 16' 34" 9.720844 9 [P. 94. r. 76.] KEY TO ALSOP'S SURVEYING. [Chap. Ill Article 143. Ex. 2. Here the half-sum of the sides is 288. "Whe ^^{Ic 167 A.C. 7.777284 214 " '' 7.669586 288 2.459392 93 1.968483 : : rad.^ 20.000000 ■ : C0S.2 1A 2)19.874745 COS. J A 30° 2' 11" 9.937372 A 60° 4' 22" Ex. 3. BC 2T.16 AB 51.6T A.C. 8.286762 AC 43.95 8.357041 2)122.78 J .9 61.39 1.788098 J-s-BC 34.23 1.534407 2)19.966308 COS. 1 A A 15° 51' 20" 31° 42' 40" 9.983154 MISCEIiliAMEOUS PROBLEMS. Problem 6. — Construction. Draw Fig. i. BC (Fig. 1) to represent the tower. At B make CBA = 98° 19'. Make BB = 5 feet. With B as a centre, and a radius of 75, describe an arc cutting BA in A. Draw AE and BE parallel to BD and AD ; then will AE be the position of the instrument. Draw AC, making DAC = 41° 18' 45", and C wdll^be the top of the tower. 10 Chap, m.] PLANE TRIGONOMETRY. [P. 114. Calculation. As sin. C 40° 22' 15'' A. C. 0.188605 : sin. BAG 41° 18' 45" 9.819653 : : AB 75 1.875061 : BC 76.44 1.873319 In this solution, AB is supposed to be equal to BE. Without this supposition the calculation would be as follows : — As EB (75) : EA (5) : : sin. EAB (98° 19') : sin. ABE (3° 46' m'^) ; And As sin. EAB : sin. AEB (77° 54' 4") : : EB : AB (74.11); Then As sin. C (40° 22' 15") : sin. BAG (41° 18' 45") :: AB :BG = 75.54. Fig. 2 Problem 7. Fig. 2. Here AGD = BDG - BAG = 14° 27' 15". Then, As sin. AGD 14° 27' 15" A.G . 0.602745 ; sin. DAG 24° 18' 9.614385 : : AD 100 2.000000 : DG 2.217130 COS. GDB 38° 45' 15" 9.892005 GB 128.57 2.109135 DG 2.217130 sin. GDB 9.796560 BG 103.2 2.013690 Height of inst. 4.75 107.95 = height required. n p. 114.] KEY TO ALSOP'S SURVEYING. Problem 8. First. To find AD. As sin. CAD 15° 51' 45'' A. C. 0.563314 : sin. ACD 103° 47' 9.987310 : : CD . 35.75 1.552276 \ : AD 2.103900 Second. To find ED. As sin. CBD 24° 7' : sin. BCD 45° 29' 30" : : CD 35.75 : BD A. C. 0.388706 9.853180 1.553276 1.795162 Third. To find ABD and AB. As BD A. C. 8.204838 : AD 2.103900 : : rad. 10.000000 : tan. x 63° 50' 22" 10.308738 And As rad. A. C. 0.000000 : tan. x - 45° 18° 50' 22" 9.533005 ABD + BAD : : tan. : tan. ABD ABJ) -BAD 64° 58' 52J" 36° 10' lOJ" 101° 9' 3" 10.330956 9.863961 As sin. ABD : sin. ADB :: AD : AB A.C. 0.008277 50° 2' 15" 9.884492 2.103900 99,236 1.996669 12 Chap. IH.] PLANE TRIGONOMETRY. Problem 9. Fig. 75. Surveying. First. To find AD. As sin. CAD : sin. ACD :: CD : AD 61° 18' 68° 15' 7.75 [P. 114. A. C. 0.056928 9.967927 0.8893Q2 0.914157 Second. To find BD and BAD. As sin. DBE 61° 20' A. C. 0.056790 : sin. DEB .75° 10' 9.985280 :: DE 7.92 0.898725 : BD 0.940795 J\D 0.914157 tan. X 46° 45' 22" 10.026638 As rad. A. C. 0.000000 : tan. x — 45° 1° 45' 22" 8.486565 BAD + ABD : : tan. 33° 37' 9.822703 BAD - ABD : tan. 2 1° 10' 4" 8.309268 BAD 34° 47' 4" Lastly. To find AB. As sin. BAD 34° 47' 4" A. C. 0.243752 : sin. BDA 112° 46' 9.964773 :: BD 0.940795 : ^B 14.10 ch. 1.149320 13 p. 116.] KEY TO ALSOP'S SURVEYING. Problem 10. The construction is as in Problem 4, page 112, Surveying. First. To find BAG. EC 4153 AC 5916 A. C. 6.227972 AB 4596 , " " 6.337620 2)14665 7332.5 3179.5 COS. 1 BAG 22° 11' V BAG 44° 22' 2'' 3.865252 3.502359 2)19.933203 9.966601 -'-^D Second. To find AE and ABE. As sin. AEG 136° 26' 30" A. G. 0.161722 : sin. AGE 19° 14' 30" 9.517926 :: AG 3.772028 : AE 3.451676 AB 4596 3.662380 tan. X 58° 23' 1" 10.210704 As rad. A.G. 0.0000000 : tan. x — 45° 13° 23' 1" 9.376451 AEB + : : tan. 2 ABE 55° 39' 29" ABE 19° 12' 2" 19.165435 AEB : tan. 2 9.541886 ABE 36° 27' 27" BAD = ABE - ADB == 17° 12' 57". Third. To find AD and BD. ' As sin. ADB 19° 14' 30" A. G. 0.482074 : sin. ABD 143° 32' 33" 9.773952 :: AE 3.662380 : AD 8287.2 3.918406 u Chap. III.] PLANE TRIGONOMETRY. [P. 115, And As sin... ABD A.C. 0.482074 : sin. B^D 1T° 12' 5T'' 9.471251 :: AB 3.662380 : BD 4127.7 3.615705 Lastly. To find DC. As sin. ADC 43° 33' 30'' A.C. 0.161723 : sin. CAD 61° 34' 59" 9.944240 :: AC 5916 3.772028 : CD 7550.8 3.877991 Fig. 5. Problem U. First. To find CA and AG. B 2::^^ As sin. CAD 21° 48' 45" A.C. 0.429959 : sin. ADC 49° 48' 30" 9.883030 :: CD 800 2.903090 : CA 3.216079 sin. ACQ 47° 22' 15" 9.866732 AG 1210.07 3.082811 Second. To find BC and BE. As sin. CBD 5° 45' A.C. 0.999184 : sin. BDC 6° 9.019235 :: CD 800 2.903090 : CB 2.921509 sin, ECB 12° 30' 9.335337 EB 180.65 2.256846 AG 1210.07 AF 1390.72 -= height required. 15 IM15,] KEY TO ALSOP'S SURVEYING. [Chap. III. Problem 12. Fig. 76. Surveying. — The measured angles being the horizontal and vertical angles, we have the follow- ing data, — AJ3EF being a horizontal plane : — AB = 252.28 feet, AEE = 80° 51' 30'^ BAE = 82° 54' 30'', ABF = 74° 37', BAF = 89° 24', and the elevations as in the problem. First. To find AE, AF, AFE, and EF. As sin. AEB 16° 14' A. C. 0.553541 : sin. ABE 80° 51' 30" 9.994448 : : AB 252.28 2.401883 : AE 2.949872 And As sin. AFB 15° 59' A.C . 0.560103 : sin. j^BF 74° 37' 9.984155 :: AB 2.401883 : AF 2.946141 AE 2.949872 tan. X 45° 14' 46" 10.003731 tan. X — 45° 0° 14' 46" 7.633005 AFE + AEF tan. 2 86° 45' 15" 4° 19' 53" 11.246335 AFE- ^EF tan. ^ 8.879340 Then As sin. AFE 91° 5' 8" A.C . 0.000078 : sin. EAF 6° 29' 30" 9.053304 :: ^E 2.949872 : EF 100.75 2.003254 Second. To find EC, EH, and CH. As rad. A.C. 0.000000 : tan. EAC 3° 45' 8.816529 : : EA 2.949872 : EC 58.40 1.766401 AP. III.] PLANE TRIGONOMETRY. [Pp. 115, 116, And As rad. A. C. 0.000000 : tan. EAH 9° 25' 9.219710 :: EA 2.9498T2 : EH 14T.7T 2.169582 EC 58.40 CH 89.37 Third. To find FD, EG, and DG. As rad. : tan. DAE 3° 54' :: EA : YD 60.22 A.C. 0.000000 8.833613 2.946141 1.779754 And As rad. : tan. GAE :: EA : EG ED DG 10° 29' 30" 163.59 60.22 103.37 A. C. 0.000000 9.267614 2.946141 CD is sensibly equal to EE = 100.75. 2.213755 Problem 13. Eig. 77. Surveying. First. To find AE and EB. As sin. AEB 42° 40' A.C. 0.168942 : sin. ABE 72° 43'- 9.979934 :: AB 7.37 0.867467 : AE 10.38 1.016343 And As sin. AEB A.C. 0.168942 : sin. BAE 64° 37' 9.955909 :: AB 0.867467 : BE 9.8247 0.992318 17 IM16.] KEY TO ALSOP'S SURVEYING. [Chap. TIT. Second. To find CE. As BE + BC 1T.774T A.C. 8.726427 : BE-BC .8747 -1.941859 BCE + : : tan. ^ BEC BEC 15° 14' 43' 37'' 9.435078 BCE- • toil 8 1 08.S6-1- . tan. ^ CJ. i.\JtJfJ\J^ BCE 15° 57' 37" BEC 14° 30' 23" And As sin. BEC 14° 30' 23" A.C. 0.601213 : sin. CBE 149° 32' 9.705040 :: BC 8.95 0.951823 : CE 1.258076 Third. To find DF and CF. As sin. DEC 27° 47' A.C. 0.331494 : sin. DCF 69° 38' 9.971964 : : DC 9.33 0.969882 : DF 18.765 1.273340 And As sin. DEC A.C. 0.331494 : sin. CDF 82° 35' 9.996351 :: DC - 0.969882 : CF 1.297727 18 :hap. Ill] PLANE TRIGONOMETRY. [P. 1]<1. Fom-tli. To find EF. CF CE 1.29772T 1.258076 tan. X 47° 36' 43'' 45° 10.039651 tan. X — 45° 2° 36' 43" 8.659140 CEF + CFE 28° 5' 19" 1° 23' 41" 9.727293 tan. ^ CFE CEF- fin 8.386433 2 CEF 29° 29' 0" CFE 26° 41' 38" As sin. CEF 29° 29' A. C. 0.307885 : sin. ECF 123° 49' 23" 9.919476 :: CF 1.297727 : EF 33.50 1.525088 AEF = AEB + BEC + CEF = 86° 39' 23". EFD = CFD + CFE = 54° 28' 38". Problem 14. First. To find AH and BH. As rad. A. C. 0.000000 : COS. BAH 3° 14' 30" 9.999304 : : AB 850 2.929419 : AH = Fa 2.928723 Fig. 6. E And As rad. : sin. BAH :: AB : BH A. C. 0.000000 8.752412 2.929419 48.07 1.681831 19 p. 116.] KEY TO ALSOP'S SURVEYING. [Chap. III.' Second. To find GE and FE. As sin. GEF 21° 27' 30" A.C. 0.436727 : sin. EFG 87° 49' 9.999685 :: FG 2.928723 : GE 2318.1 3.365135 And As sin. GEF A.C. 0.436727 . : sin. FGE 70° 43' 30" 9.974947 :: FG 2.928723 : FE 2189.8 3.340397 Third. To find FD, GD, AF, and BG. As sin. FDG 44° 32' A. C. 0.154081 : sin. FGD 47° 39' 9.868670 :: FG 2.928723 : FD 894.3 2.951474 And As sin. FDG A. C. 0.154081 : sin. DFG 87° 49' 9.999685 :: FG 2.928723 : DG 1209.2 3.082489 DF 2.951474 tan. A DP 3° 25' 45" 8.777584 AF 53.59 1.729058 BH 48.07 BG 101.66 Fourth. To find CK and CE. AK 3.340397 tan. CAK 35° 27' 9.852466 CK 1559.1 3.192863 AF 53.59 CE 1612.69 20 CHAPTER IV. CHAIN SURVEYING. Ex. 2. 17.25 + 16.43 + 14.65 Article 251. 48.33 2 j- sum j- sum sum sum -17.25 -16.43 -14.65 2 24.165 6.915 7.735 9.515 = 24.165 = half sum of sides. 1.383187 0.839792 0.888460 0.978409 2 )^4.089848 Area, 110.9 eh. = 11 A., E., 14.4 P. 2.044924 Ex. 3. 19.58 + 16.92 + 12.76 -19.58 -16.92 -12.76 2 24.63 5.05 7.71 11.87 49.26 -— - = 24.63 = 1 5. 2 ^ 1.391464 0.703291 0.887054 1.074451 106.69 ch. = 10 A., 2 R., 27.04 P. 2)4.056260 2.028130 Article 254. ' Ex. 2. Fig. 107. Surveying. 2AED =AE.ED = 4.15 x 8.25 = 34.2375 2EFCD = EF.(ED + CF) = 5.02 x 17.61 = 88.4022 2 CFB = FB.FC = 2.26 x 9.36 = 21.1536 2) 143.7933 71.89665 ch. 7 A. R., 30.3 P. 21 Pp. 146, 140.] KEY TO ALSOP'S SURVEYING. [Chap. IV. Fig. 7 Ex. 3. A E B F 2AEC =AE.EC = 3.76 x 5.68= 21.3568 2 CEFD = EF . (CE + FD) = 9.89 x 14.60 = 144.3940 165.7508 2BDF =BF.FD = 3.98 x 8.92= 35.5016 2) 130.2492 65.1246 ch. = 6 A., 2 R., 1.99 P. Article 265. Fio-. 8. First field. Bases. Perpen- diculars. Areas. 109.8900 102.4320 2ABCF 2FCDE ' 9.90 9.70 11.10 10.56 2 )212.3220 10 A., 2 R., 18.576 P. = 106.161 ch. Second field. 2EKGD 13.15 8.75 115.0625 2e:ghi 12.85 7.20 92.5200 2GIHL 11.50 4.32 49.6800 2) 257.2625 12 A., 3 B., 18.1 P. = 128.63125 ch. 22 Chap. IV.} CHAIN SURVEYING. [Pp. 149,151. Fig. 9. First field. Areas. Bases. Perpen- diculai'S. Areas. 2 ABC ' 2ACDE 2AEFG 11.50 12.55 10.51 4.82 12.65 9.20 55.4300 158.75T5 96.6920 2 )310.8795 15 A., 2 E., 7 P. = 155.43975 ch. Second field. 2FaHL 2HIKL 13.95 10.80 13.00 10.85 181.3500 117.1800 2 )298.5300 14 A., 3 E., 28.24 P. = 149.265 ch. Ex. 2. First tract. Article 256. Dis- tances. Offsets. Inter. Dist. Sum of Offsets. Double Areas. 0.00 0.00 Left baud 7.90 6.00 7.90 6.00 47.4000 areas. 19.20 4.00 11.30 10.00 113.0000 21.45 3.75 2.25 7.75 17.4375 31.27 0.00 0.00 9.82 3.75 36.8250 0.00 2.00 4.65 2.00 4.65 9.3000 Eisjlit liand 8.40 5.16 6.40 9.81 62.7840 areas. 14.85 5.23 6.45 10.39 67.0155 20.70 6.42 5.85 11.65 68.1525 25.90 4.76 5.20 11.18 58.1360 31.27 0.00 5.37 4.76 25.5612 25 A., 1 E., 4.89 P. 2 )505.6117 = 252.80585 ch. 23 Pp. 151, 157.] KEY TO ALSOP'S SURVEYING. Second tract. [Chap. IV. Dis- tances. Offsets. Inter. Dist. Sum of Offsets. Double Areas. 0.00 0.00 1.75 2.37 1.75 2.37 4.1475 10.26 7.29 8.51 9.66 82.2066 21.37 4.35 11.11 11.64 129.3204 28.46 5.94 7.09 10.29 72.9561 40.25 0.00 11.79 5.94 70.0326 0.00 0.00 4.29 6.23 4.29 6.23 26.7267 15.48 3.19 11.19 9.42 105.4098 32.54 8.26 17.06 11.45 195.3370 36.17 7.92 3.63 16.18 58.7334 40.25 0.00 4.08 7.92 32.3136 2 )777.1837 38 A., 3 K., 17.47 P. - 388.59185 ch. Article 257. Ex. 2. First. To find ABC. 21.69 + 23.72 + 14.50 AB BC AC ABC 29.955 • 8.265 6.235 15.455 154.457 1.476469 0.917243 0.794836 1.189068 2)4.377616 2.188808 24 Chap. IV.] CHAIN SURVEYING. [P. 157. Second. To find ABE. 21.69 + 13.96 + 13.44 49.09 ^ , 2 2 — ^d-:r.,tjnLKJ, 1 9 2 "^ 24.545 1.389963 Is - AB 2.855 0.455606 J-s - BE 10.585 1.024691 J 5 - ^E 11.105 1.045518 2)3.915778 ABE 90.759 1.957889 = 26.22. 1.418633 0.397940 0.945469 1.173186 2 )3.935228 CBD 92.814 1.967614 25 To find CBD. 23.72 + 17.40 + 11.32 : 52.44 2 2 1 9 2 "^ 26.22 is- CB 2.50 is- BD 8.82 is - CD 14.90 V. I'jT.] Offsets. KEY TO ALSOP'S SURVEYING, [Chap. IV Base. Distance. Offsets. Int. Dist. Sum of Offsets. Double Areas. 000 4.80 .75 4.80 .75 3.6000 CD 7.39 .97 2.59 1.72 4.4548 8.96 .72 1.57 1.69 2.6533 10.54 .55 1.58 1.27 2.0066 11.32 .00 .78 .55 .4290 5.51 6.52 .50 1.01 .50 .5050 8.24 .45 1.72 .95 1.6340 FB 10.16 .87 1.92 1.32 2.5344 12.37 .35 2.21 1.22 2.6962 14.14 .63 1.77 .98 1.7346 17.40 3.26 .63 2.0538 24.3017 .1 00 2.95 .75 2.95 .75 2.2125 DF 4.52 .75 1.57 1.50 2.3550 5.51 .99 .75 .7425 Subtractive ABC ABE CBD 5.3100 2)18.9917 9.4958 154.457 90.759 92.814 Area, 34 A., 3 R., = 347.5258 eh. 2G CHAPTER V. COMPASS SURVEYING. Article 326. Ex. 1. Fig. 127. Surveying. To find BD, AD, BAD, and ABD. As sin. DBE 37° hU A. C. 0.211468 : sin. DEB 120° 45' 9.934199 : : DE 18 1.255273 : DB 1.400940 As sin. Di\E 46° 35' A. C. 0.138839 : sin. AED 61° 20' 9.943210 : : DE 1.255273 : AD 1.337322 BD 1.400940 tan. X 49° 10' 54" 4° 10' 54" 10.063618 tan. X — 45° 8.863999 BAD + ABD tan. ^ 64° 37' 30" 8° 45' 47" 10.323947 BAD - ABD tan. ^ 9.187946 BAD 73° 23' 17" ABD 55° 51' 43" = = DGC. DGF = 180° - 55° 51' 43" = 124 g/ Yl". To find DC, DG, and CG. As sin. DCE 46° 35' A. C. 0.138839 : sin. DEC 26° 50' 9.654558 : : DE 1.255273 : DC 1.048670 27 p. 188.J KEY TO ALSOP'S SURVEYING. [Chap. V. As sin. DGC 55° 51' 43" A. C. 0.082134 : sin. DCG 38° 53' 17" 9.797821 :: DC 1.048670 : DG 8.48 0.928625 As sin. DGC A. C. 0.082134 : sin. CDG 85° 15' 9.998506 :: DC 1.048670 : CG 13.47 1.129310 Ex. 2. Fig. 120. Surveying. To find BC, CA, CAB, and ABC. As sin. CBD 42° 28' A. C. 0.170593 : sin. CDB 47° 29' 9.867515 : : DC 7.00 0.845098 : CB 0.883206 As sin. CAD 47° 18' A. C. 0.133763 : sin. ADC 100° 25' 9.992783 : : CD 0.845098 : AC 0.971644 CB 0.883206 tan. X 50° 47' 5° 47' 38" 38" 10.088438 tan. X — 45° 9.006330 ABC + BAC tan. 2 61° 7' 10° 25' 20" 10.258336 ABC - BAC tan. ^ 9.264666 ABC 71° 32' 20" BAC 50° 41', 40" ACF = BFC - BAC = 16° 53' 20". DCF = DCA + ACF = 49° 10' 20". 28 Chap. V.] COMPASS SURVEYING. [P. 188. To find AF and FC. As sin. AFC : sin. ACF : : AC : AF And as sin. AFC : sin. CAF : : AC : CF 67° 35' 16° 53' 20'' 2.94 50° 41' 40" 7.84 A. C. 0.034124 9.463171 0.971644 0.468939 A. C. 0.034124 9.888617 0.971644 0.894385 Fig. 11. Ex. 3. Construction. — Take AB' any length. At A and B' make the angles B'AE, B'AF, AB'E, and AB'F equal to the given angles. Join EF: produce it to a, making EG = 10.78. Through G draw GD parallel to AE, meeting AF in D. Draw DB parallel to FB', and BC parallel to B'E. Then will AB he the line and CD the known base. By Art. 316 we have AD : AC : : sin. ABD . sin. ACB : sin. ADB . sin. ABC whence As AD or : AC or : : rad. : tan. x tan. 45° — X ACD + ADC sin. ABD 80° 47' sin. ACB 60° 54' sin. ADB 52° 43' sin. ABC 37° 23' A. C. 0.005643 '' " 0.058602 9.900722 9.783292 10.000000 tan. tan. ACD ACD - ADC 29° 15' 10" 15° 44' 50" 72° 23' 30" 41° 37' 11" 114° 0'41" 9.748259 9.450213 10.498422 9.948635 29 p. 189.] KEY TO ALSOr'S SURVEYING. [Chap. V As And (Art. 316) sin. CAD ^ sin. ADD j sin. ACD \ sin. ADB CD AB 35° 13' 80° 47' 114° 0'41" 52° 43' 10.78 13.76 Ex. 4. To find BD, AD, and ABD. A. C. 0.239073 " " 0.005643 9.960691 9.900722 1.032619 1.138748 Fig. 12. As sin. CBD 54° 25' A. C. 0.089765 : sin. BCD 33° 18' 9.739590 : : CD 15 1.176091 : BD 1.005446 As sin. CAD 36° 2' A. C. 0.230434 : sin. ACD 103° 10' 9.988430 : : CD 1.176091 • : AD 1.394955 BD - 1.005446 tan. X 67° 48' 22° 48' 45" 45" 10.389509 tan. X — 45° 9.623888 ABD + BAD tan. ^ 64° 15' 41° 6'' 30" 10.316805 ABD - BAD tan. ^ 9.940693 ABD 105° 21' 30" GFD = 360° - (FDB + DBG + BGF) = 72° 21' 30". FPD = GFD - DDF = 33° 31' 30". 80 Thap. v.] compass surveying. [P. 189. To find PD and FD. As sin. CPD 35° 40' A. C. 0.234280 : sin. PCD 105° 30' 9.983911 : : CD 15 1.176091 : PD 1.394282 And as sin. PFD 72° 21' 30" A. C. 0.020921 : sin. FPD 33° 31' 30" 9.742175 : : PD 1.394282 : FD 14.37 1.157378 Ex. 5. Fig. 118. Surveying. To find CA, CB, CAB, and CBA. As sin. CAD 41° 15' A. C. 0.180887 : sin. CDA 43° 45' 9.839800 : : CD 10 1.000000 : CA 1.020687 As sin. CBD 55° 1' A. C. 0.086547 : sin. CDB 87° 39' 9.999635 : : CD 1.000000 : CB 1.086182 CA 1.020687 tan. X 49° 18' 15" 4° 18' 15" 10.065495 tan. X — 45° 8.876584 CAB + ABC tan. ^ 61° 10' 7° 47' 6" 10.259233 C^B -- ABC tan. 2 9.135817 CAB 68° 57' 6" ABC 53° 22' 54" 81 p. 189.] KEY TO ALSOP'S SURVEYING To find CE, and AE, and DCE. As sin. AEC : sin. CAE : : AC : CE 5.192 [Chap. V- 81° 43' 29° 19' 54'' A. C. 0.004554 9.690076 1.020687 0.715317 And as sin. AEC : sin. ACE : : AC : AE 68° 57' 6" A. C. 0.004554 9.970011 1.020687 9.89 0.995252 DCE = DCA + ACE = 95° + 68° 57' 6" = 163° 57' 6". Ex. 6. To find DC. As sin. DCE 45° 56' A. C. 0.143554 : sin. DEC 67° 25' 9.965353 ^ : : DE 9.25 0.966142 : DC 11.89 1.075049 ■ Fig. 13. A;2 4 Bearings. Ch. Bear. Dist. 16.55 11.48 N. 12.68 S. E. W. N. 47° W. N. 40° W. 10.64 N. 19° 5' W. N. 12° 5' W. 11.23 2.40 N. W. 15.53 (14.45) (5.69) N. 2.3° E. N. 30° E. 9.72 14.00 8.42 4.86 13.87 5 6 N. 75i° E. N. 82i° E. 1.89 S. 7° E. North. (48.67) 18.73 18.73 41 p. 219.] KEY TO ALSOP'S SURVEYING. [Chap. V. As Dist. : Dep. : : rad. 15.53 5.69 sin. ch. bear. K 21° 30' W. Bearing, S. E. K 28° 30' W. A. C. 8.808829 0.755112 10.000000 9.563941 As rad. : COS. ch. bear. : : Dist. : Dep. 14.45 A.C. 0.000000 9.968678 1.191171 1.159849 Article 352. Ex. 2. Sta. 1 2 3 4 5 Bearings. Ch. Bear. Dist. N. S. E. W. S. 29i° B. S. 9i° W. 3.19 3.15 .53 5.70 S. 37i° AY. S. 761° W. 5.86 1.37 S. 39i° E. South. (11.28) (11.28) N. 53° E. S. 871° B. 19.33 .76 19.31 N. 31° 5' W. N. 8° 10' E. (16.26) (16.10) (2.31) 7.01 6 7 S. 60i° w. N. 80° W. 7.12 1.24 2.15 S. 29i° E. S. 10° W. 2.18 8.12 .38 j 8 S. 60i° W. N. 80i° W. 1.37 8.00 1 18.71 18.71 21.62 21.62 As sin. ch. bear. 8° 10' : rad. : : Dep. 2.31 : Dist. 16.26 A. C. 0.847549 10.000000 0.363612 1.211161 As rad. : COS. ch. bear. : : Dist. : Lat. 16.10 A.C. 0.000000 9.995573 1 .211161 1.206734 42 Chap. V.] COMPASS SURVEYING. [P. 221. Article 353. Sta. 1 2 4 6 7 8 Bearings. Dist. N. s. E. w. S. 29f ° E. 3.19 2.77 1.58 S. 37J° W. 5.86 4.66 3.55 F. 53° E. 19.32 11.63 15.43 S. 60f ° W. 7.12 2.18, 3.48 1.07 6.21 S. 291° E. 1.90 S. 601° w. 8.12 4.00 7.07 11.63 16.81 18.08 11.63 16.83 Lat. and Dep. of closing Hue 5.18 1.25 16.83 As lat. clos. line : Dep. : : rad. : tan. bear. 5.18 1.25 K 13° 34' W. A.C . 9.285670 0.096910 10.000000 9.382580 As COS. bear. : rad. : : Lat. : dist. clos. line 5.33 A. C. 0.012290 10.000000 0.714330 0.726620 Clos. line 1 ACB ACB AB 5.33 BC 16.26 AC 11.29 2)32.88 16.44 11.11 4° 4'. 8° 8'. Fisr. 14. A. C. 8.788879 " " 8.947306 1.215902 1.045714 2) 19.997801 9.998900 43 SOP'S SURVEYING. [Chap. V. A.C. 9.273380 1.052694 9.150686 ;' 9.476760 Bear, of BC = 17° 26' + 13° 34' = 31° 0' S.E. 5th side. Bear, of CA = 31° + 8° 8' = 39° 8' 'N.W. 3d side. Pp. 221, 226.] KEY TO AI And As AB 5.33 : AC 11.29 : : sin. ACB : sin. ABC 17° 2< ' Article 357. Ex. 2. Here ABC = 37° 15'. As rad. A. C. 0.000000 : sin. ABC 37° 15' 9.781966 TAB '' IBC 17.25 1.236789 10.87 1.036230 : ABC 113.497 2.054985 ABC 56.7485 ch. = 5 A. , 2 R., 27.98 P. Ex. 3. As rad. A.C. 0.000000 : sin. A 126° 47' 9.903581 .. JAB •• lAC 23.56 16.42 1.372175 1.215373 : 2 ABC 309.834 2.491129 ABC 154.917 ch. = 15 A. Article 358. , 1 R., 38.67 P. Ex. 2. As / ''^'^• ^' I sin. C A. C. 0.000000 42° 21' " " 0.171561 ( sin. A * tsin. B 63° 52' 73° 47' 9.953166 9.982367 TAB •' lAB 17.63 1.246252 1.246252 : 2 ABC 397.7385 2.599598 ABC 198.8692 ch. = 19 A., 3 R., 21.9 P. 44 Chap. V.] COMPASS SURVEYING. [Pp. 225, 227. Ex. 3. As{^"^- { { sm. sin. sin. : double area Area 57° 31' 63° 17' 59° 12' 15.65 222.771 111.3855 ch. A. C. 0.000000 0.073890 9.950968 9.933973 1.194514 1.194514 2.347859 11 A., K., 22.17 P. Article 359. Ex. 2. Here ABC = 93° and BCD = 106°. As rad. A. C. 0.000000 : sin. ABC 93° 9.999404 ( AB 12.47 1.095866 •• \BC 11.43 1.058046 2.153316 A. C. 0.000000 9.982842 1.058046 0.961985 2.002873 A. C. 0.000000 9.512642 1.095866 0.961985 : 1st quantity 142.336 As rad. : sin. C /BC '' ICD 106° 9.16 : 2d quantity 100.664 As rad. : sin.B + C- /AB '''' I CD 180° 19° : 3d quantity 37.196 100.664 142.336 2 ABCD = ABCD = 280.196 140.098 ch. 1.570493 = 14 A., R., 1.56 P. 45 p. 227.] KEY TO ALSOP'S SURVEYING. [Chap. V, Ex. 3. Here B = 48° 30' and C = 120° 30'. As rad. : sin. B f AB ••* IBC : 1st quantity 48° 30' 8.63 9.2T 59.917 A.C. 0.000000 9.874456 0.936011 0.967080 1.777547 As rad. : sin. C TBC : 2d quantity 120° 30' 11.23 89.697 A.C. 0.000000 9.935320 0.967080 1.050380 1.952780 Area = As rad. : sin. 180° - (B H- C) 11° TAB '• ICD : 3d quantity 18.492 59.917 + 89.697 •- 18.492 A. C. 0.000000 9.280599 0.936011 1.050380 1.266990 131.122 = 65.566 eh. = 6 A., 2E., 9P. 46 Chap. V.] COMPASS SURVEYING. [P. 235. CO CO O I— I CO o w O $ o S o -^ o 02 O O to o to co" o CM o oo oo o 1 in 125 1 CO I— ( t^ o -^" o £ •l-l o o o o r-H CO .1—1 OS Co' cq CO CM cvi CO I— 1 o I— 1 1 1 1 — 1 CO 1—1 2 --J OC 1 (M O ^ CO O GO ■ ^ to 1 CO 00 ^ CO -^ T— 1 '^ CO o C5 O O 1 CO Ci 1 rH CO O ^ -* 1 O O -«:f 0-3 Cs ^ O ci CO '^ G5 1 » CO 1 1 o 1 1 1>- 1 a6 1 I— 1 rH ^-1 O O rH 1 1 1 o O o o o 1— 1 ^ o oo o to f4 CO I— 1 CO !— J o Oi CO W2 O o o r— ( CO co' o , 12^ 00 00 S 1>^ o o 1— 1 O lO CO as oo CO CO 9 f4 V o T— 1 o V lO (M o CO lO to to to CM lO CO o CO M 1— 1 (M 1 CO Tf to CO to o o CO I— 1 1 — 1 ^ r- t^ C5 O ^ CO 00 i-H I— 1 rH to si cd CO OS OS to' p -* CO rH CO §1 i ^ 00 CO CO* rH 1'—' 10 CO r— 1 B 10 as I— ( as IS CO !>; rH tzi CO to ^* to CO 0* 1—1 b- rH 1—1 CO CO Z t— 1 I— ( 00 OS to (M* I— 1 i^ ! ^ 0^ 1—1 T-{ ^ 1 r-l I 10 CO 1 ca 1 CO 1 T-H 1 1 C4 rH 1 1 1 |,-l CO t-H 1 "^ ^ 1-H 1 1 t- 1 CO I OS 1 b-. j CO* rH 1 -T-{ I— ( 1 DQ 1 t- 10 CO 1 '^ 1 1 1 rH CD rH I— ( CO !>; CO* CO ^H 1 CO 00 1 UO ^ 2^ "^ ! Sj . ( r-H 10 to GO -g rj" -:^ 10 00 p oc CO cm' 1 t-^ 1 CO CO CO* I-H OS CO to c 03 '^ CO -cf V CO »o CO to to to It- s to V to 00 CO CO 1 rH o lO o CO -t^ oo CM o o o o CO* CM CM oo lO CM O J.-I <^^ I CO ^ oo o cq I ^ I ^ '-0 CO no T— I o I ^ CO CO "^ h» 12; \^ t^ 02 fci U^ 1^ ^ hzi cq O I t^ 00 00 o OO CX; <» CO CM T— 1 r— 1 CO t- CO Oi CO o oo o CM Ci C75 O O O lO OO oo CO oo CM lO o lO CM 49 p. 235.] KEY TO ALSOP'S SURVEYING. [Chap. V. X to 2 •< m o O CO CO I— J o 1—1 o >o Ci CO o to 1—1 oi CO 00 Ol r-H o" o 04 Ol Ol Ol CO Ol CO S3 O O >o o o to to -^* CO i-i .2 oo' r— 1 o CO 1—1 oo o Co' 00 to oq o oq to' CO oq CO • oo o CO 1—1 o CO 1—1 (M od .— 1 o oo 1—1 to oq T— i to o6 CO o o oq 1—1 oo Ol I— 1 oq to to* 1—1 ^ 00 CO oo r— 1 oq o oc* H lO GO o 1— 1 T— 1 oq I— J o o 1 — 1 to lO* QQ o oq 1—1 CO oq CO I— 1 ^ I— 1 • 1—1 Co" 1—1 r— 1 00* o oo '^' CO 1— ( I— 1 1—1 CO" I— 1 cm' I— 1 r— 1 C5 oq oq co' T— 1 uo o ^' o oo o ' 1—1 oq oq o CO •^* 1—1 CO o 00 .S O o 1—1 to V O 1—1 o I— 1 o o 00 o V to I— 1 o O CO m o to o CO to CO o 1^ 00 c ■ o N o 'J2 o -^ o "A V O CM o "it) CO 3 V 1-.0 oq Ol 00 m 3 o oo V O CO o 'A V o Ol o CO CO V to to o 1—1 o V to —1 o oq to 1 (M CO 1 -^ O o \r~ 00 05 o 1 — 1 to ^ S5 <==> Ci -* a. oc (M O CO CO to r-l -^ 1^ Ci 1—1 o to Ol o li ^ o CO CO O Ttl -^ 1— i 1— ( 1~— 1-1 :X to . o oq o I-- oq -^ o o o o o o Ol to o o CO t— o oo O Ci O Oi O oo r^ ^■i CO C-3 CO o o Ci CO o o Ci CO C-5 CO to (-1 oo o o Ci "^. o i^ o CO oq P^ cq to '*' u r— < * -1^ • ^ 3D 33 -+J to o 00 Ci o o o* to Jf^ od Oi O TJ (/i rC! T^ P ^ -4-3 05 QQ r-l r3 'TS 60 Chap. V.] COMPASS SURVEYING. [P. 235. Tl2 < GC O CO •CO o -^ Ci o o "^ ci 1—1 GO -^ o to' Ci < oo o oo CI o CI o r— 1 o UO oo oo o CO* CO o CO o GO d GO o -^ CI (M o CO O Ci t^ -*_ CO iC -*' OO CI rH CI j ^ CO ci I— 1 Ci* cq r— 1 O Ci oo CI -*. o O ^ CO CI .— 1 r-H o '2. — ^ t^ 1^ O O o o oo o I— 1 czi o Ci 1—1 1 1 1 1 CI 1 CO Tti lO o t— oo Ci o o cq l-l -^t* TJH T^l o xf t^ oc lO 1^ CO -t^ r— ' ■oc Ci t- ^ to cl r—^ GO -* cf Ci I— 1 rH a; C-1 I— 1 ' — ■ . Ph c^ ^ o o o o o o Cl Oi CO .-H CO <-> '^ CI C^i 1—1 r-H CO Ci r-H cq - CO ^ a CO a:> oo ,^ ^ cJ o to .»H I-H fH a 'Ci 00 .. CO oo <1 •• • ■ o -^ t^ o O O Ci <-Ti o GO o CO rH -^ CO o c.O o to o o CI o o o .— 1 1^ o o •CO CO o o o t— 1 i-H* l—J o* « '^J^ -( CO CO r— 1 r-H Ol (M GO CM CM i < ! \ 1 is* CO 1— 1 p4 QO 1 '"' p4 CO CXD I t- CO 1 -* r-H 0* rH (M 1 d 1 CO t- CM 1 1 Tf CO CO rH •CO 1 1>^ rH 1 w CO I-^ 1 l-H 1 CM* ,-1 1 1 1'^ OJ 1 1 T-^ a' - 1 1 1 '-' 1 ^ rH -* (M' r-H i—t T^ 1^ ^ (M rH 1 rH rH 1 * 1 ^' CO CO • CO r-K T-H w i-H c i 1 CO 00 1 T-H l-H 1^ 1 (M s CO I— ' I— 1 CO Co' rH ^. CO* rH 10 bo C 1 ' l^ 1 CM CO "A CO [» - j CO ■^ lO -* t- CO to rt ^ rH , G<1 00 Cj -^ r~ 0-2 f>J 'O r-\ CO uo i- c:. Tt^ CO CO r^ . cq 1.58 1.27 2.0066 -a .78 .55 .4290 < 1.01 .50 .5050 1.72 .95 1.6340 1.92 1.32 2.5344 2.21 1.22 2.6962 1 1.77 .98 1.7346 3.26 .63 2.0538 j 24.3017 2.95 1.57 .99 .75 1.50 .75 2.2125 2.3550 .7425 _5.3100 is".99iT 52 Chap. V.] COMPASS SURVEYING. [P. 23^ o o t^ CO O t— o i—i O CO -l '^ J^- 00 cq o ai Si oo O CO (M ; t-; -3 -*' ai oi (M g (M (M r-^ 1 W.D.D. CO 1— ( CO t^ 1:^ OS oi '^' ^' T-i CM P P O O Oi iq cji c^i to f4 1— 1 (M ^ 1^ a lO ,—1 ^ 1>^ T— 1 1>^ > I— ^ as --^ ;-i (M o O . P3 W c4 Ir-^ '^' CO GQ I— 1 CD >o Ci CO 02 lO 1— ( o CO rH - 1— I 1—1 00 ■7^ I— 1 (M O CP (M 1—1 ^. •iH • i—i ;2q o6 cc OO* •^ < &>-■ o'^ c4 rH M •^ ir- P^ ^' to t—i t-^ t-^ I— 1 rH Ci- -H CN co OO «■ c4 1—1 '*' CD Oi to rH ai 1—5 1— 1 00 T— 1 CM O CN rH "*. ^ cx). cx5 co' 1 o o C-J 1— 1 O aj o 'Ci ^. o to P r-^ o od o6 CJ I CO T^ CO o O CO '^ •^ Tf to (M 1— O 00 to -h CO CM . p^ <1 5a o o <1 o ^ on to b- o 1—1 cq - 1— ( QO CO C<1 CO I— 1 OS p4 h h CO OS CO CO CO lO ^ 1-J c4 —1 C oo o -^ CO O o i^ . lO O CO rH O O CO ■^ CO o OS CfJ T-l ?t5 « -^ <5 •• .. OS o o OS ^ lO CO C ^ ^ '^ o 1—1 I— 1 CO 1-1 o o I— 1 ->* CO CO 00 o CO o r-l ""^l bo n © £_( a ci Ol ""* r^ CD o o . 285-288.] ] lEY T ALSOP'S SUR\ ^ey; [NG. [Chap. VII. Article 399. Ex. 2. As AEC 58 ch. A. C. 8.236572 : ADE 30 " 0.477121 (AVi 12.25 1.088136 '' JAC 10.42 1.017868 : AD.AE 1.819697 AE 8.50 0.929419 AD 7.77 . .890278 Ex. 2. Article 400. As 5 : 2 : : AC^ (400) : AD2 = 160, .-.AD = ^/ 160 = 12.65. Article 401. Here, CF bein [g perpendicular to AK , we have As rad. A. C. 0.000000 : COS. A. 39° 45' 9.885837 : : AC 13.24 1.121888 : AF 1.007725 AG AD 6.365 8.049 .803798 2)1.811523 .905761 ' 68 Chap. VII.] LAYING OUT AND DIVIDING LAND. [P. 290. Article 402. Ex. 2. To find AE and EP. . As sin. AEP 111° 30' : sin. APE 33° 45' :: AP 5.90 : AE 3.523 And As sin. AEP : sin. Pi\E 34° 45' : : AP : PE 3.614 To find AG. As AE : AF 5.935 :: AD 4.4 : Aa = KI + IG = 7.412 2EP 7.228 El-IG .184 KG = 1.168 Fig. 18. A. C. 0.031322 9.744739 0.770852 0.546913 A. C. 0.031322 9.755872 0.770852 0.558046 A. C. 9.453087 0.773421 0.643453 0.869961 - 1.264818 2 )0.134779 0.067389 And AK = AG ± KG = 7.412 ± 1.168 = 6.244 or 8.58. 69 Pp. 291,293.] KEY TO ALSOP'S SURVEYING. [Chap. VII. Article 403. Ex. 2 Fig. 19. DF = 1 CD = 3.96 and AE = i AB = 4.97T. To find PM and AM. As sin. M 83° 50' A. C. 0.002520 Fig. L : sin. PAM 54° 50' 9.912477 M V : : AP 6.20 0.792392 A '\' i : PM 5.098 0.707389 H \\iL. As sin. M A.C . 0.002520 E --^'^. : sin. APM 41° 20' 9.819832 : : AP 0.792392 : AM 4.119 0.614744 ^ D find ML, LP, and FG. -• B ^ As AD 5.18 A.C. 9.285670 : AM 0.614744 : : AE - DF 1.017 0,007321 : AE-ML AE .809 . 4.977 — ■1.907735 ML 4.168 LP = MP -ML = .93 As AM — J AD 1.529 : JAD 2.59 : : PL .93 : FG = EH 1.575 A. C. 9.815593 0.413300 -1.968483 0.197376 whence DG = DF + FG = 5.535, and AH = AE - EH = 3.402. Ex. 2. EF=J( Article 404. 4BC2 + 5 . ADS / /426.0096 + 154.0125^ 9 ■) = J0 9 = v/ 64.4469 = 8.03. 70 Chap. VII.] LAYING OUT AND DIVIDING LAND. [Pp. 293, 294. And as BC - AD (4.77) : FE - AD (2.48) : : AB (6.47) : AE = 3.36. Ex. 2. To find AG and BG. Article 405. As sin. G : sin. B : : AB : AG As sin. G ; sin. A : : AB : GB 34° 62° 30' 19.55 31.01 83° 30' 34.74 A. C. 0.252438 9.947929 1.291147 1.491514 H A. C. 0.252438 c 9.997199 1.291147 1.540784 Fiff. 20. To find GH and BI. AsGA 31.01 A. C. 8.508486 : GD 18.76 1.273233 : : GO 15.82 1.199206 : GH 9.57 1.980925 BI = f (GB - GH) = 13.98 ; GI = GB - BI = 20.76. To find GF and BF. AsGP 25.01 A.( 1 8.601886 : GA 31.01 1.491514 : : GI 20.76 1.317227 : GF 25.74 1.410627 GB 34.74 BF 9.00 71 p. 297.] KEY TO ALSOP'S SURVEYING. [CHAr. VII. Article 406. Ex. 2. As in Ex. 2 of last b ^ article, find AG = 31.01, GB = 34.74, GD = 18.76, GC = 15.82. Hence, AsGB : GC : : GD : GH 34.74 15.82 18.76 8.544 A. C. 8.459216 1.199206 1.273233 0.931655 AH = GA - GH = 22.466 AI = f AH = 12.481, GI = 18.529, and GK = 9.264. To find BO, OP, and GL. As sin. BOP 34° A. C. 0.252438 : sin. BPO 126° 15' 9.906575 :: BP 11.52 0.061452 : BO 16.614 1.220465 As sin. BOP A. C. 0.252438 : sin. PBO 19° 45' 9.528810 : : BP 1.061452 : PO 6.961 0.842700 As GO ■ 18.126 A. C. 8.741698 : GB 34.74 ' 1.540784 :: GK 9.264 0.966799 : GL 17.753 1.249281 72 Chap VII.] LAYING OUT AND DIVIDING LAND. To find EL and AE. [P. 301. E]^-hL^ = LG 1.249281 Fl^ - LIST = LG - -2P0= 3.831 0.583312 2)1.832593 EL = 8.247 .916296 AE = AG - (EL + LG) = 5.01. Article 407. Ex. 2. r sin. E 77° A. C. 0.011276 128° 45' u " 0.107970 J sin. A 76° 9.986904 * [sin. D 78° 15' 9.990803 (AD 8.35 0.921686 '' ' [AD a 0.921686 : Eourthterm 87.162 1.940325 174.324 3BC2 119.07 5)293.394 EE= x/ 58.6788 = 7.66. And As sin. G 25° 45' A. C. 0.362065 : sin. E 9.988724 : : EE - BC L36 0.133539 : BE 3.05 .484328 AE = AB - BE = 1.60. Ex. 3. * f sin. E Ass . -ci t sin. E 77° A. C. 0.011276 78° 15' (( " 0.009197 f sin. A • t sin. B 76° 128° 45' 9.986904 9.892030 r AB • • I ^B 4.65 4.65 0.667453 0.667453 : zy^ 17.152 1.234313 ■r. 302.] KEY TO ALSOP'S SURVEYING. [Chap. VII. I /^X2f + ^CJ)\ // 51.456 + 106.58x ^ ^^ -- - W V 5 J~^\ 5 ^~"-" And As sin. G- 24° 45' A. C. 0.378139 : sin.'E 77° 9.988724 : : CD - EF 1.68 0.225309 : FA 3.91 Article 408. 0.592172 , r sin. E Asi • 17, t sm. F 65"^ 15' A. C. 0.041846 90° '' " 0.000000 / sin. A • I sin. B 76° 128° 45' 9.986904 9.892030 r AB • * I AB 4.65 4.65 0.667453 0.667453 : xy^ 18.017 1.255686 , f sin. E As i . -c. I sm. F A. C. 0.041846 " " 0.000000 f sin. C • t sin. D 77° 78° 15' 9.988724 9.990803 f CD • = t CD 7.30 0.863323 7.30 0.863323 : vw"^ 55.978 1.748019 EF-J(^^^^ -1- 2 ?'?r^\ ^ ^ ^"^ ) = ^ 33.2014 = 5.762. 5 ^ As sin. E A. C. 0.041846 : sin. B 9.892030 : : AB 4.65 0.667453 : BO 3.993 0.601329 And As sin. G 24° 45' A. C. 0.378139 : sin. E 9.958154 : : EF - BO 1.769 0.247728 : AF 3.837 0.584021 CHAPTER VIII. MISCELLANEOUS EXAMPLES. Ex. 1. Make AB = 50, one ^^s- 22. of tlie given sides; erect the ,'"' ""^-^. ,. , ^^ 2 Area perpendicular BE = AB 19.2. Ty^itli the circle A, and radius equal to the other given ~jy side, (32,) describe an arc cut- ting EC, a parallel through E, in C and C^* then will ABC or ABC be the triangle required. Calculation. — AD = s/AC^ My = v/ 655.36 = 25.6, and (12.13.2) BC^ = BA^ + AC^ ± 2 BA . AD = 2500 + 1024 ± 2560 = 6084, or 964 ; whence BC = 78, or 31.05. Ex. 2. Let ABCD be the o;arden. Produce AB, makins: BI = I AB. On AI describe the semicircle ALI, cutting CB produced in L. Make AE = AD, and bisect EB in M. With the centre M and radius ML describe a semi- circle cutting AB in F and K. Make BH = BK: then will AF or CH be the breadth of the walk. I K /D p. 303.] KEY TO ALSOP'S SURVEYING. [Chap. VIII. For FH = FB . BK = BP = AB . BI = I AB . BC = I ABCD ; also, MF = MK, and EM = MB ; whence FE = BK = BH; .•.AF = HC. Calculation.— BI = | BC = 87.5 ; BL^ = AB . BI = 10500. ML = x/MB2 + ]3L2 ^ ^10600 = 102.95. AF = AE + EM - MF = 7.05. Ex. 3. Similar triangles are as the squares of their homologous sides. The area of the triangle, whose sides are 3, 4, and 5, is 6. Whence 6 ch. : 24 ch. : : 16 : 64, and n/64 = 8, one side; 6 and 10 ch. are the others. Ex. 4. Construction. — Let ABC be the plat. Produce the diameter CA to E, making AE = i AD. On EC describe a semicircle cutting the perpendicular DB in F. DF will be the radius of the outer circle. For, since circles are as the squares of their radii, we have ABC : IFG : : DC^ : DFl But CD : DF : : DF : DE; whence CD^ : DF^ : : CD : DE : : 4 : 5 ; .-. ABC : IFG : : 4 : 5. Calculation.—^ : 5 : : AD^ : DP ; : 5625 : 7031.25, and DI = 83.85; whence AI = 8.85. 76 Chap. VIIL] MISCELLANEOUS EXAMPLES. [P. 303. Ex. 5. Let AB represent the tree, and C and D the two stations. Then we have ACE = a 5° ST', ADE = 2° 29', BCE = 67° 43' 30'', CD = 100, and CA = 75.. As sin. CED 3° 8' A. C. 1.262332 : sin. CDE 2° 29' 8.636776 : : CD 100° 2.000000 : CE 79.27 ' 1.899108 tan. BCE 67° 43' 30" 10.387619 BE 193.52 2.286727 BE^ 37451 4.573454 As CE + CA 154.27 A. C. 7.811719 : CE-CA 4.27 0.630428 CAE : : tan. + 2 2 CEA 87° 11' 30" CEA 29° 26' 3" 11.309328 CAE : tan. 9.751475 CEA 57° 45' 27" As sin. CEA 57° 45' 27" A. C. 0.072733 : sin. ACE 5° 37' 8.990660 :: CA 75 1.875061 : AE 8.679 0.938454 AE^ 75.32 1.876908 EB2 37451 AB = v/ 37526.32 = 193.7. 77 p. 303.] KEY TO ALSOP'S SURVEYING. [Chap. VIII. Ex. 6. Describe the triangle ADE, having the sides ED, AD, and DE equal to 6.23, 4.95, and 5.62 respect- ivel3\ On DE describe the equilateral e^ triangle DBE. Join AB, and on it de- scribe the equilateral triangle ABC. This will be the one required. For, since ABC = EBD, ABE = CBD ; whence, in the two triangles ABE and CBD, we have two sides and the included angle of one equal to two sides and the included angle of the other; consequentl}^ AE = DC. Calculation, AD 4.95 DE 6.23 A. C. 9.205512 AE 5.62 " " 9.250264 2); 16.80 8.40 0.924279 3.45 0.537819 2)19.917874 cos. 1 AED 24° 31' 29'' 9.958937 i\ED 49° 2'' 58" As EB + EA 11.85 A. C. 8.926282 : EB - EA .61 -1.785330 EAB + : : tan. ^ EBA 35° 28' EBA 2° 6' 31" 3" 9.852871 EAB - : tan. ~ 8.564483 Then As sin. EBA : sin. AEB :: EA : AB 33° 22' 28" A.C. 0.259541 109° 2' 58" 9.975540 5.62 0.749736 0.984817 Chap. YIIL] MISCELLANEOUS EZiAMPLES. [P. 304. Finally, 5 As rad. A. C. 0.000000 : sin. 60° 9.93T531 r AC 0.984817 •• IaB 0.984817 : 2 ABC 80.754 1.907165 ABC = 40.377 cli. = 4 A., E., 5.6 P. Fis:. 27. Ex. 7. As sin. AEC 7° 43' A. C. 0.872007 : sin. ACE 44° 34' 9.846175 : : AC 60 1.778151 : AE 2.496333 As rad. A. C. 0.000000 : sin. EAF' 37° 43' 9.786579 :: AE . 2.496333 : EF 191.83 2.282912 _30 Height 221.83 And As rad. A. C. 0.000000 : cos. A 9.898201 : : AE 2.496333 : AF 248.05 2.394534 p. 304.] KEY TO ALSOP'S SURVEYING. [Chap. VII Ex. 8. Through A conceive the hori- zontal plane AEF to pass, and draw BG parallel to EF. Then we have FAE = 87° 45', FAC = 39° 43', FAD = 52° 13', BAE = 2° 17', AEF = 54° 23', GBC = 33° 4', and GBD = 45° 42'. First. To find AE. As rad. : COS. BAE 2° 17' : : AB 157 : AE 156.88 A.C. 0.000000 9.999655 2.195900 2.195555 To find AF, FC, and FD. As sin. AFE 37° 52' A. C. 0.211955 : sin. AEF 54° 23' 9.910054 :: AE 2.195555 : AF 208.54 2.317564 tan. CAF 39° 43' 9.919448 Elev. ofhill = FO 172.5 2.237012 As rad. - A. C. 0.000000 : tan. FAD 52° 13' 10.110579 :: AF 2.317564 : FD 268 2.428143 FC .172.5 Height of tower CD = 95.5 The calculation might have been made by finding BE = 6.25, EF = BG = 255.37, GC = 166.26, and GD = 261.69. Whence FC = 172.51, FD = 267.93, and CD = 95.43. 80 Chap. YIII.] MISCELLANEOUS EXAMPLES. [P. 304. Ex. 9. It is readily seen that AC : BC : : tan. DHC : tan. DFC : : cot. DFC : cot. DHC. But AC : BC : : sin. ABC : sin. BAC. ^^Oience D Fig. 29. - f ^\ A ] As cot. DHC 32° A.C. -1.795789 : cot. DFC 39° 48' 10.079267 : : sin. BAC 90° 10.000000 : sin. ABC 48° 35' 23'' 9.875056 As sin. BCA 41° 24' 3T" A.C ). 0.179506 : sin. ABC 9.875056 :: J\B 100 2.000000 : AC 2.054562 tan. 39° 48' 9.920733 94.47 1.975295 10 CD 84.47 Ex. 10. Make BD = 9.43, BDE = 49° 50', DBE = 72° 41', DA = 7.56, Fig. 30. and AE = BE BD X 8.42 : also make ABC will = EBD and BAG = BED : then ABC be the triangle required. For, since ABC and EBD by con- struction are similar, we have EB : BD : : AB : BC. But ABC = EBD ; whence ABE = CBD. Consequently (6,6) ABE 6 81 p. 304.] KEY TO ALSOP'S SCIIVEYING, [Chap. YIIl. and CBD are similar ; whence BE : EA : : BD : DO, or BE BE AE = — - . DC. But by construction AE = x 8.42. BD -^ BD Whence DC = 8.42 ; and, since AD = 7.56 and BD = 9.43, the conditions of the problem are fulfilled. To find ED. As sin. BED 57° 29' A. C. 0.074051 : sin. DBE 72° 41' 9.979855 :: DB 9.43 0.974512 : DE 10.6762 1.028418 To find AE we have by construction BD : BE : : DC AE; or, As sin. BED : sin. BDE :: DC cos. 1 ADE ADE whence ADB 57° 29' 49° 50' 8.42 22° 48' 29" 45° 36' 58" ADE + EDB A.C. 0.074051 9.883191 0.925312 : AE 7.6305 0.882554 To find ADE. AE 7.6305 AD 7.56 A.C. 9.121478 DE 10.6762 8.971582 2)25.8667 is 12.93335 1.111711 Js - AE 5.30285 0.724510 2)19.929281 9.964640 = 95° 26' 58". CHAP.Vni.] MISCELLANEOUS EXAMPLES. [P To find AB. As BD + DA 16.99 A.C . 8.769807 : BD - DA 1.87 0.271842 BAD 4- : : tan. — ABD ABD 42° 16' 31" 5° 42' 51'' 9.958631 BAD - : tan. — 9.000280 As sin. BAD 47° 59' 22" 0.128998 : sin. BDA ^ 95° 26' 58" 9.998032 :: BD 9.43 0.974512 : BA 12.63 1.101542 [P. 304 To find BC and AC. And As sin. C • sin. BAC : ; AB • BC As sin. C J sin. ABC \ * AB : AC 49° 50' 57° 29' 13.94 72° 41' 15.78 A. C. 0.116809 9.925949 1.101542 1.144300 A. C. 0.116809 9.979855 1.101542 1.198206 Ex. 11. Let AB = 50. Bi- sect it in D, and makeDF = 42.5 the half sum of the two sides. At D erect the perpendicular DG. Make DE = 2 Area -4- AB = 32. Lay off" BG = DF. With D as a centre, describe the arcs FI and GH, the latter ^ cutting EC parallel to AB in H. Join DH, and jDi'oduce it to I. Draw ICL perpendicular to AB, cutting EC in C. Then will C be the vertex of the triangle. It is evident, from the construction, that the fease and p. 304.] KEY TO ALSOP'S SURVEYING. [Chap. VIII. area agree with the data. It remains to be proved that AC + BC = 85. Put AD = DB = a, DF = BG = 6, DG = c, DL = d, and LC = e: then we have a^ -f c^ = 6^; or 6^ — c^ = a^ By similar triangles, DP : DH^ : : IL^ : LC^; or b^ : c^ : : h'' — d^ : &\ whence & = — -— = & — - — . AC^ = AL^ + LC^ = (a + c^)^ + 6^ = a^ + 2 «c? + 6^2 + c=^ - —^h^ + 2ad+ ^^ =b^ + 2ad + — ; ad whence AC = b -{■ -~ Similarly BC = 6 - y .-. AC + BC = 2 6 = 85. Calculation. DG^ = DF^ - DB^ = 1806.25 - 625 = 1181.25. Then, As DH^ : DP : : DE^ : IL^ or As 1181.25 : 1806.25 :: 1024 : IL^ = 1565.7989; whence DL = v/DP - IL^ = 15.5065, and AC = >/ AL^H- LC^ = n/ 1640.77654225 + 1024 = 51.6215; ... BC = 85 - AC = 33.3785. Ex. 12. Make AB = 20 Fig. 32. the given side, and at A erect the perpendicular AF ^/ = 2ABC-r.AB = 7. Di- ''"" vide AB in E, so that AE :EB::3:2. Take BG ^ ^^" ' a third proportional to AE — EB and EB. With the centre G and radius GE describe a circle cutting FC parallel to AB in Q and C^ Then will ABC or ABC be the tri- angle required. 84 CiiAP.VIIL] MISCELLANEOUS EXAMPLES. [P. 304. For, since AE - EB : EB : : EB : BG, by composition, AE : EBi.-EG : BG. Alternately, AE : EG::EB : BG; whence, by composition, AG : EG : : EG : BG. and (F.6) AC : CB, or A'C : C^B : : AE : EB : : 3 : 2. Calculation. AE = 12, BE = 8, BG = 16, GE = 24, DC = 7; whence GD = GD' = v/GC^- CD^ = ^527 = 22.96. Then (12.13.2) BC^ = BG^ + GC^ ± 2 BG . GD = 256 + 576 ± 734.72 = 97.28, or 1566.72. Whence BC = 9.86 or 39.58, and AC = i BC = 14.79 or 59.37. 85 p. 305.] KEY TO ALSOP'S SURA^EYING. [Chap. Vlll. 1—1 m OS •< O 1— 1 o I— 1 1-5 I— ( OO CO as 1— 1 OO I— 1 O >o CM )>; Co" I— 1 Si .2 is* o o I— 1 o CO 3 o Ci co' o I—I I— 1 o o o o CO o CO --1 1—1 o CO co' ^ I— 1 s I— 1 to oq CO 1— 1 CO ai oo OO I— J 1^ oo CO s CO CM i-H co' » 1 fcO o m ^ &; V V O o I— 1 CO o o Oi -* OO I— ( 5ta Cl, o ^' 6 Ph -t-> 80 Chap. VIII.] Ex. 14. MISCELLANEOUS EXAMPLES. [P. 30- Bearings. S. 26i° E. K 60J° E. K 281° W. Dist. N. s. E. w. j 1.53 1.37 .68 ■ 1 i 19.37 9.54 16.86 1 3.27 2.87 1.56 (11.04) (15.98) 1 As diff. lat. : dep. : : rad. : tan. bear. As COS. bear. : rad. : : diff. lat. : dist. 12.41 12.41 17.54 17.54 11.04 15.98 K 55° 22^ E. ^b"" 22' 19.43 A. C. 8.957031 1.203577 10.000000 10.160608 A. C. 0.245405 10.000000 1.042969 1.288374 Bearings. Dist. N. s. E. w. S. 55° 22' W. 8.53 4.85 7.02 , S. 261° E. 1.53 1.37 .68 K60i° E. 8.53 4.20 7.42 (2.02) (1.08) As diff. lat. : dep. : : rad. : tan. bear. As COS. bear. : rad. : : lat. ; dist. 6.22 6.22 8.10 8.10 2.02 1.08 K 28° 8' W. A. C. 0.694649 0.033424 10.000000 9.728073 A. C. 0.054604 10.000000 0.305351 2.29 .359955 87 V. 305.] KEY TO ALSOP'S SURVEYING. [Chap. VIII. Ex. 15. A slight investigation ^ig- 33. will show that the distances from the observers to the point directly under the balloon are as the co- tangents of the angles of elevation. These cotangents are, a = .73412, h = .94345, and c = .75996. Hence the following construction : — Let A, B, and C be the places of the observers. Take AE = AB and AF = AC. At the points E and F, c b with radii equal to - . AE and - . AF, a a describe arcs cutting in G: join AG and make ACD = AGE. Then will D be the point under the balloon. Since the two arcs will cut in G' as well as G, there will be two points (D and D') that will answer the conditions. Because AGE = ACD, the triangles ADC and AEG are similar ; whence AD : DC : : AE : EG : : a : c; and AG : AE : : AC : AD .-. AG : AC (AF) : : AE . (AB) : AD. Since, therefore, AFG and ADB have the common angle at A and the containing sides proportional, they are similar, and consequently AD : DB :: AF : FG : : a : 6. Calculation. To find FG and EG As a .73412 A. C. 10.134230 : c .75996 -1.880790 ' :: AE 1000 3.000000 : EG 1035.19 3.015020 As a A. C. 10.134230 : b .94345 -1.974719 : : AF ' .1690 3.227887 : FG 2171.88 3.336836 88 Chap. VIIL] M ISCELLANEOUS EXAMI >LES. [P. To find AEF an id FEa. AF 1690 AE 1000 . A. C. 7.000000 EF 1180 2)3870- " " 6.928118 1935 3.286681 i^- AF 245 2.389166 2)19.603965 cos J AEF 50° 39' 57" 9.801982 AEF 101° 19' 54'' Fa 2171.88 ^ FE 1180 A. C. 6.928118 EG 1035.19 2)4387.07 " " 6.984980 2193.535 3.341145 is- Fa 21.655 1.335558 2)18.589801 COS. J FEa 78° 37' 37" 9.294900 FEG 157° 15' 14" AEC = 360° - (AEF + FEa) = 101° 24' 52" = ADC. To find DAC and DCA. As AD (a) - A. C. 10.134230 : DC (c) -1.880790 : : rad. 10.000000 : tan. X 45° 59' 26" 10.015020 As rad. A.C . 0.000000 : tan. X - 45° 59' 26" 8.237799 : : tan. DAC + ACD 2 DAC - ACD 2 39° 17' 34" 48' 38" 9.912902 : tan. 8.150701 DAC 40° 6' 12" 89 p. 305.] KEY TO ALSOP'S SURVEYING. [Chap. VIII. 101^ 24^52'' A.C. 0.008677 40° 6' 12'' 9.808999 1690 3.227887 1110.6 3.045563 52° 46' 10.119210 1461.4 3.164773 or AEG' = FEG - AEF = 55° 55' 20" = AD'C. Then, as AD' : D'C : : rad. : tan. x. = 45° 59' 26" as before. And As rad. A. C. 0.000000 As sin. ADC J sin. DAC • • AC. : CD tan . He ight : tan. x — 45° 59' 26" 8.237799 D'AC : : tan. + ACD' 2 -ACD' 2 62° 2' 10" 1° 52' 0" 10.275037 D'AC : tan. 8.512836 D'AC 63° 54' 20" As sin. AD'C 5."^° 55' 20" A.C. 0.081825 : sin. D'AC 63° 54' 20" 9.953311 :: AC 1690 3.227887 : CD 1832 3.263023 tan. 52° 46' 10.119210 Height 2411 3.382233 Ex. 16. ' To find the area of the whole tract. Sta. 1 2 N. s. E. w. E.D.D. W.D.D. Multipliers. S. Areas. 10.82 1 19.78 57.71 15.81 2.79 22.57 22.57 W. 356.8317 3 G.99 12.48 9.69 12.88 W. 90.0312 ! 4 5 6 7 6.75 15.64 28.12 15.24 E. 102.8700 1.56 22.35 37.99 53.23 E. 83.0388 16.26 10.03 32.38 85.61 E. 1392.0186 9.05 37.93 27.90 57.71 E. 522.2755 ; 1 33.62 33.62 60.50 60.50 90 2 ) 2547.0 058 1273.5329 Chap. VIII.] MISCELLANEOUS EXAMPLES. [P. 305. To find the area to a line north, from the first station. 1 sta. 1 2 3 4 N. s. E. w. E.D.D. W.D.D. 19.78 Multipliers. S. Areas. 10.82 19.78 ■ 1 15.81 2.79 22.57 22.57 W. 356.8317 6.99 12.48 9.69 12.88 W. 90.0312 4.35 (10.09) 22.57 9.69 E. 42.1515 5 (29.27) 10.09 19.78 E. 578.9606 33.62 33.62 22.57 22.57 1067.9750 As rad. : cot. bear. 4th side S. 66° 40' E. :: Dep. 10.09 : Lat. ■ 4.35 S. A.C. 0.000000 9.634838 1.003891 0.638729 1273.5329 - 1067.9750 = 205.5579 = double area between the division line and a line due north 29.27 ch. from the first station. Then As 1 ^!°- t sm. 76° 35' A.C. 0.012017 66"^ 40' " " 0.037055 ( rad. \ sin. 36° 45' 10.000000 9.776937 : : 2 area 205.5579 2.312934 : fourth term 137.703 2.138943 (29.27)^ = 856.7329 v/719.0299 = 26.815 = division li As sin. 36° 45' A.C. 0.223063 : sin. GQ"" 40' 9.962945 1 1 2.455 0.390051 : Dist. from 1st cor. 3.77 0.576059 Ex. 17. As it is of no importance to the calculation of this question what the course of the line is, we shall sup- 91 p. 306.] KEY TO ALSOP'S SURVEYING. [Chap. VIII. pose it to be north. The bearings of the difierent lines in course will then be as below. Sta. Beariugs. Dist. N. 3.81 s. E. w. 1 2 3 4 5 South. (23.67) 9.24 K 67° 35' E. 10.00 K 18° 52' E. 7.25 6.86 2.34 K 7° 7'E. 5.43 5.39 .67 K 58° 10' W. (14.42) (7.61) (12.25) As rad. A.C. 0.000000 : cot. bear. 5th side 58° 10' 9.792974 : : Dep. 12.25 1.088136 : Lat. T.61 0.881110 As sin. bear. A.C . 0.070793 : rad. - \ 10.000000 : : Dep. 1.088136 : Dist. 5th side 14.42 1.158929 Distance of line, 23.67 . Fig. 34. Ex. 18. Here we have AE : CE : : cot. CAE : rad., and DE : EC : : cot. ODE : rad. .-. AE : ED : : cot. CAE : cot. CDE. Whence cot. CDE : cot. CAE : : sin. DAE : sin.' ADE. As cot. CDE 20° 4' A.C. 9.562636 : cot. CAE 25° 10' 10.328037 : : sin. DAE 90° 10.000000 : sin. ADE 51° 1'40" 9.890673 92 Chap. YIIL] MISCELLANEOUS EXAMPLES. [P. 306. As rad. A. C. 0.000000 : tan. ABE 10.092061 : : AB 60 1.778151 : AE 74.17 1.870212 As rad. A. C. 0.000000 tan. CAE 25° 10' 9.671963 : : AE - 1.870212 : EC 1.542175 As EC A. C. 8.457825 : EB 4.17 0.620136 : ; rad. 10.000000 : cot. EBC 83° 10' 34'' 9.077961 Ascos. EBC A. C. 0.925118 : rad. 10.000000 : : BE 0.620136 BC 35.1 yds. 1.545254 93 p. 306.] KEY TO ALSOP'S SURVEYING. [Chap. VIII. Ci X ^ o CO I I o o o o • o O CO C7i CJ t-~ CO rH ^ t^; CJQ m ^ ^ H* p4 W V V V O lO O (M -* ^ o o o o .-^ O O O CO O rH CO O OO '^. ^ !z; 'n 02 1 c.t -Tf UO rH O lO CM O CO 1-. o ci CO CM t— (M CM OO I- o OO '^ CM 00 OO Oi o t- O CO CM CO* O J o r-T -TtH ^ t- o t^ -^ o o 1—1 d J—{ OC Q c3 a> < u c3 © rQ g o T-A CO ft o o b^ CO I— I CM c— ) ■^ CM CO -^ (-) CM t^ CO (— ) rH CM '^ O t^ a> i-H o T-H o T-H O -< t^ OO CM '-J CM iO ^ W V o .— 1 r— 1 o o O Oi fO >o <_) r- on o OC , I— 1 rH ;2; CO OQ d rd o CO (M V Oi U3 125 OQ o p ^^ d .2 lO ^ O -^ 02 © .S c3 © © g 94 CHA.P. VIIL] MISCELLANEOUS EXAMPLES. rp. 306. Ex. 20. Let ABCJ) be Fig. 35. the quadrilateral. Divide ^ it into two equal parts by the line EF parallel to CD. ^- \Q (Art. 407.) Then divide the quadrilaterals ABCD ^^ ,^ ^j^ and AEEF into two equal ----- \n parts by the lines GH and g:.^— :.____^^_^_-_._ _^ii IK parallel to AD. Join "^^s^^-z^"':^ HK, and produce it to M. ' '' \ p ' Through G draw GL pa- a— — -^ ^ rallel to MH. Then will ML be the other division line. For, since GL is parallel to MH, the triangles MGH and MLH are equal : to each add AMHD, and we have AMLD = AGHD = lABCD. Again, since XK is parallel to LH and FN" to GL, we have MX : XL : : IsIK : KH : : MI : IG. Whence IX is parallel to GL. The triangle MIK is therefore equal to MXK : to each add AMKF, and we have AMXF = AFKF = 1 ABCD. To find AF and EF. * f sin. C ' As<^ . -p. ( sm. D 78° 45' A. C. 0.008426 63° " " 0.050119 ' sin. A 78° 30' 9.991193 ( sin. B 139° 45' 9.810316 f AB 23 1.361728 '•'' \AB 1.361728 : fourth terra 383.274 2.583510 CD^ 2161.3201 2)2544.5941 EF x/ 1272.2970 = 35.67. 95 306.] KEY TO ALSOP'S SURVEYING. [Chap. VTII As sin. 38° 15' A.C. 0.208243 : sin. C 78° 45' 9.991574 : : CD. - EF 10.82 1.034227 : FD 17.14 1.234044 AD 49.64 AF 32.50 To find AGr and GH. fsin. A ^' I sin. D A.C. 0.008807 a (( 0.050119 r sin. B * I sin. C 9.810316 9.991574 /BC '' IBC 30.50 . 1.484300 1.484300 : fourth term 675.15 2.829416 AD^ 2464.1296 2^ (3139.2796 GH = ^/1569.6398 = 39.62. As sin. 38° 30' A.C. , 0.205850 : sin. D 9.949881 : : AD - GH 10.02 1.000868 : AG 14.34 1.156599 Draw EQ parallel to AD. Then As sin. CEQ 38° 15' A.C. , 0.208243 : sin. CQE 9.949881 :: CQ "10.82 1.034227 : EC 15.57 1.192351 BC 30.50 BE 14.93 96 Chap. VIII.] MISC ELLANEC To find AI and IK , As / ''''• ^ ' ^^ \ sin. E r sin. B • tsin. E (BE '' IBE 14.93 : fourth term 161.79 AE^ = 1056.25 [P. 306. A. C. 0.008807 " " 0.050119 9.810316 9.991574 1.174060 1.174060 2.208936 2)1218.04 IK = v^609.02 = 24.68. As sin. 38° 3 : sin/E :: AE - IK 7.82 : AI 11.19 A. C. 0.205850 9.949881 0.893207 1.048938 ^Now, because IK is parallel to GH, we have GH : IK : : MG : MI; whence, by division, aH-IK:IK::ia:MI; or 14.94 : 24.68 : : 3.15 : MI = 5.20 ; and AM = AI - IM = 5.99. Through M and H draw MO and HP parallel to AD and AB. Then in HPO we have As sin. HOP (D) A. C. 0.050119 : sin. HPO (A) 9.991193 : : HP 8.35 0.921686 : HO 9.18 0.962998 As sin. HOP A. C. 0.050119 : sin. PHO 38° 30' 9.794150 : : HP 0.921686 : HO 5.83 0.765955 And by similar triangles MOH and GHL we have MO (45.45) : OH (9.18) : : GH (39.62) : IIL = 8. 7 97 p. 306.] KEY TO ALSOP'S SURVEYING. [Chap. VIII. Then in MOL we have MO, OL, and MOL, to find OML. Thus 8.203218 1.451326 10.212681 9.867225 As MO + OL 62.63 : MO - OL 28.2T MLO + OML ::tan. ^ 58° 30' MLO - OML 36° 23' . tan. ^ OML 22° 7' Bearing of MO S. 66^ 15' E. " " ML S. 88° 22' E. 98 THE END. BTETJSOTYPED BY L. JOHNSOX & CO. PUILADEU?HIA. m'^' 6 A 8 0) Deacidified using the Bookkeeper process. Neutralizing agent: Magnesium Oxide Treatment Date; Jan. 2004 L PreservationTechnologJes * A WORLD LEADER IN PAPER PRESERVATION j 1 1 1 Thomson Park Dnve . Cranberry Townshin. pa isnfiR i LIBRARY OF CONGRESS