TRIGONOMETRIC FORMULAE 
Solution of Oblique Triangles 
Given Required . . .• jj , . sy 
£ k U V= 180*-(a +*).■« 6 ’ n 
A , B t a b, c, O 
A , a, b 
<*> ft. 
«> ft, « 
<*» ft, « 
A. ft, «? 
-4, 2?, 
A* 
A, JJ.d 
A, B C 
Area 
Area 
Area 
sin B — 
b sin A 
•in A 
,< 7 = 180 ° — (A -t-B),e = ln 
* ' Vj ■ aid .4 
4 + -0=180°— 0. u» l (A—B)= ^ a ~ b ^ * a - i ^ 
a 4- ft 
a sin G 
e N 
n A 
hs—'b'Hn—e) 
*= 2 ' 
.in y]^^. C=m°-4A + B ) 
Q -j- h *4- C / — ! ■ . ■ - - . 
« = , area = V •(• — a) ( 9 — b) (*— c) 
area — 
ft c sin A 
area = 
a 1 gin B sin O 
REDUCTION TO HORIZONTAL 
Horizontal distance— Slope distance multiplied by the 
cosine of the vertical ancle. Thus: slope distance 319.4 fL 
Vert angle — 5° ltK. From Table. Page IX. cos 6" W — 
J| .9959. Horizontal distance— 319. 4>: .9950^318.09 ft. 
r: Horizontal distance also — Slope distance minus slope 
" distance times (1— cosine of vertical angle). With the 
same figures as in the preceding example, the follow- 
ing result is Obtained. Cosine 5°10' .9959. 1-9959= 0041. 
919 4X 0041 = 1.31. 319.4-1.31 =318.09 ft 
When the ri.^e is known, the horizontal distance is approximately:— the s 
n* el- the stiuaro of the rise divided by twice the si op exists nee. Thus: rise -14 ft. 
slope distance=302.B ft llortzontal distance^ 302.6- -_>U± -302. 6-0 32- 302. 28 ft. 
2X802: 6 
Horizontal distance 
MADE >tt i>. 6< A. 
