B. H. Hill, I Site levels June 1934–Dec. 1935Bert Hodge Hill

Drawing: diagram with points C, D

Margin left: 44

Margin left: 1/2

C 117 D

CDE: 38°20' 117. v=2 CD 118°40' + 23°00'

Drawing: diagram with points C, D, E

C 117 D 86.74 E 139.30 86.74

Drawing: diagram with points C, D, E

C 185.74 E 139.30 E 123.44

392.25 384.96 C 117 D 262.74 185.74 E

G

Arrow

CDG solution p 48, foot. 117 = log 2.06819 392.25 = " 2.59356 384.96 = " 2.58542 262.74 = " 2.41952 185.74 = " 2.26887

Drawing: diagram with points D, E, F, G, f

Labels: 20.55, 122°5' 1/3, E, G, 27.12' 1/3, 326.90, F, f, 25°28', 48°30'

D

CDE 38°20, 84°53, 56°47 CD = 117 m

DEE 35°47', 123°13', 23°0' DE 185.74

392.25 G CGE 39°30', 17°17', 123°13' CG = 392.25

Above [triangle] solved by law of sines

E GE

Drawing: triangle with points E, G

Labels: E, 56°47', 123.44, 298.22, G

298.22 123.44 - 174.78 + 421.66 123°13 / 2 = 61° 36' 30" Solve by law of tangents ∠ E = 99° 5' 20" & G = 24° 7' 40" EG = 252.62

[triangle] DGF DG = 384.96 ∠ F = 73.58 hence FG = 326.90 & DF = 312.74 [arrow] [triangle] DEF gives DF = 312.71

Right page

Takings ∠ s D G & F as observed 54.42 + 51.20 + 74.5 = 180° 7' Must of course be reduced to 180°

Solutions [triangle]s DEF & GEF with a single outer angle reduced by 7' (D 20°48', DFE 25°53', EFG 47.58 or G 27°5' 1/3) show provi all impossible & so the correction must be made partly in each [triangle]. This can be best done at F, where the divis in 26° + 48°5' is in any case inaccurate

Assuming then ∠ s D & G correct, first [triangle]s DFE & GFE. Produce GE to meet DF at f. Soln [triangle] DEf: DE = 185.74, ∠ DEf = 57° 54' 2/3 & ∠ D = 20.55 ∠ DfE = 101° 10' 1/3' (hence ∠ EfF = 78°49' 2/3) Df = 160.39 (hence fF =) & fE = 67.59 & so fEG = 320.21 (252.62 + 67.59)

[triangle] GfF give fG = 320.21 & ∠ s = 27° 12' 1/3, 78.49' 2/3 & (..) 73° 58' fF = 152.32 fG = 326.85 by law of sine

in [triangle] EfF give sides fE 67.59 & fF 152.32 & ∠ EfF = 78°49' 2/3 solve by law of tangents = ∠ fFE = 25°28' 4 1/2 ∠ fEF = 75°42' 15 1/2" & EF = 154.22

[triangle] AGEF EG = 252.62 ∠ F = 48° 30' hence EF = 154.23 (73.58 - 25.28) & FG = 326.92