Notebook·PU-Mu 1050, Box 8, folder 5·June 1934–December 1935

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

Text·

Handwritten in English. Every line is transcribed, and the text sits beside the scan.

Scan 29 of notebook PU-Mu 1050, Box 8, folder 5 (B. H. Hill, I Site levels June 1934–Dec. 1935), handwritten in English. The transcription follows.

Scan: Penn Museum Archives, Philadelphia, PU-Mu 1050, Box 8, folder 5, Courtesy of the Penn Museum.

Unnumbered pageScan 29 of 87

[58]	In △ ACD if side AC = 201.74
	CD = 117
Solve ~~for C~~ assuming ∠ A = 25.56 ; ∠ D = 48.47 1/2

	A) Sin 25.56	9.64080	D) Sin 48.47 1/2	9.87640
	Ls 117	2.06819	Ls 201.74	2.30481
		7.57261		7.57159
	" 201.74	2.30481	117	2.06819
	104.4	9.87742	131.1 1/2	9.63978
	Sin D	48° 56 1/2	S. A	25° 52'
	C.	105. 7 1/4		105.20 1/2

~~assuming ∠ C = 105.22~~	201.74
~~∠ D + A = 74.38~~	117
~~2 = 37-19~~	d - c	84.74
	d - c	1.92809	192809	d + c	318.74
	d + a	2.50343	S	25.10.5
		9.42466
	32.99 Tan	988210	Sin	9.78763
		9.30676		1.71072	side
	D - A	11° 25' 21	Sin	9.79675	A - D
	D + A	37 - 19		2.41397	= 259.40
	D	48.44.21
	A	25 - 53 - 39	105° 20'
			D + A = 74.40
		9.42466	37.20
		9.88236
	Ta	9.30702	25.10.5	1.92809	62
		11° 27' 46"		9.78280
		37.20		1.71089
		48.47.46	+ Si	9.29825
		25 - 52.14		2.41264
				= 258.60

				3° 45
		259	310
	Lg 310 =	2.49136	In △ ACD let
	Si 3°45	8.81560	reduce ∠ D to
		1.30696	(= 20.275)	say 48.40 or 48.42
	259	2.41330	Obser ∠ 171.37.30
		8.89366	∠ A then
		= ∠ 4° 30' 23	∠ D = 171 - 40 - 37	25 - 50

Right page

△ ABD	[inserted: p 57] Side BD = 242.52 (Lg 238475)
∠ A = 69.43	∠ B = 110.15 - 21.13 = 89.2	[inserted: (15?)]
	∠ D = 48.47 - 27.23 = 21.24	[inserted: 21.4]

	Ls 242.52	238475		2.41255	(drawing)
	Ls 69.43	9.97220			(drawing), "D 48.47]
		2.41255	S 88.53	9.99997
	Ls 21.24	9.56215		2.41249
1)		1.97470		= 258.50
		94 35
	of 94.28	p 56 foot	7. p 58	+ p 56
		2.41255	S	2.41255
2)	S 21.15	9.55923	S 89.2	9.99994
	93.71	1.97178	93.71	2.41249	258.52
		2.38475	assuming Ls B + D correct
	S. 69.34	9.97178		2.41249	+ A = 69 34
		2.41297	S 89.2	9.99994
3)	21.24	9.56215		2.41291
		= 1.97512	94.43	= 258.77

If in △ BCD we reduce ∠ C by 9' = 131.° 15 B + D = 48.45 = 24° 22 1/2'

2

2

	Lg T =	9.65619	S =	9.61504
foot of p 57 =		1.50133		1.50133
		1.15752		1.11697
	24.22.30	2.42442		2.38450
	3.5.46	8.73310	- S	8.73247
D = 27.28.16	= 3° 5' 46"		2.38450
B = 21-16.44			= 242.38

The in △ ABD we have BD = 242.38 ∠ B = (110-15 - 21.17 =) 88.58+ + D = (48.47 - 27.28 =) 21.19

		2.38450	Ls C B D A as taken
	S 69.43	9.97220	= 131.24
		2.41230	110.15
4)	21.19	9.56053 =	A. B.	48.47 1/2
		1.97283 =	93.94	69.43
	88.58	9.99993	A. D.	360°.9 1/2
		2.41223 =	258.36	must be reduced
				by 9 1/2'

65 figures hard to read~~struck~~ struck outadded written in above” jug ditto, and what it repeatsword reading not certain

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