TRIGONOMETRIC FORMULAS 
B B 
B 
b_ 
c 
a 
T» cosec = — 
ft a 
Given 
a , ft 
a . c 
A , a 
A, 6 
-4, c 
Given 
^4, 77, a 
^4, a, ft 
a , ft, C 
Required 
A, B ,c 
A, 77, ft 
B , ft, c 
77, a, o 
7?, «, ft 
1 — 
v 1 «* 
a , ft, c 
a, ft, c 
^4, ft, c 
^4, 1?, 6', a 
Required 
ft, c, (7 
77, 6*, G 
A, B, c 
tan -4. = — = cot B, c — \/ a 2 + ft 2 — « 1 + “^r 
sin J. = — = cos 17, ft = \/ (c + a) (c—a) = c 
c 
77 = 90° — A, ft = a cot A, c = a 
77 = 90° — A, a = ft tan A, c = , 
cos -4 
77 = 90° — A, (i — c sin A , ft = c cos -4 
Solution of Oblique Triangles 
a sin 77 ^ a sin G 
ft = — 
sin A 
ft 
A, B, G 
Area 
Area 
Area 
sin 
sin A 
_ a sin G 
sin A 
G = 180 ° — (A + B),c = 
sin B = h SU1 A ,C = 180 ° — (A -t-£),e 
(l 
-l+iJ-180 0 — C, tan 1 ( A—B )- "* tan * { , ‘ l il ) ~ 
' ».i„c ,, + s 
c = 7 
sin A. 
s= *±6±C s . nU= ^-6Xs-c) 
2 
ft c 
Sin 1 B = ( 7 = iso 0 — (*4 + #) 
\ AC 
s = 
®4-ft-f-C /— 7 — Tw \ 
, area = v $($ — «) (a — ft) (* — c) 
La 
ft c sin A 
area = 
area = 
a 2 sin 77 sin (7 
‘7 sin y4 
REDUCTION TO HORIZONTAL 
Horizontal distance 
Horizontal distance — Slope distance multiplied by the 
cosine of the vertical angle. Thus: slope distance =319. 4 ft. 
Vert, angle — 5° 10'. Since cos 5° 10' =.9959, horizontal 
distance=319.4X. 9959=318.09 ft. 
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. 
319.4X. 0041 = 1.31. 319.4-1.31=318.09 ft. 
When the rise is known, the horizontal distance is approximately the slope dist- 
ance less the square of the rise divided by twice the slope distance. Thus: rise=14 ft., 
slope distance=302.6 ft. Horizontal distance=302.6— 14 * ** =302.6—0.32=302.28 ft. 
2X 302.6 
