•te, * 
TRIGONOMETRIC formula 
B 
n 
a 
c 
Right Triangle 
b' ' C 
Oblique Triarigles ' 
0 
Solution of Right Triangles 
F6r Angle A, sin - cos =' — , tan = ^ , cot = ~ , sec = 7, cosec «= — 
- o a (i 
* * * 
Given 
a, b . 
Required 
\/ a 2 -f b* = 
it 
sin A = 7 =cosB t b = \/{o+a)(c — a) = d ^1 — ^ 
/?.=^90° — 4, b = a cot -4, <5 = 
* = *tan 4,4 
COS il. 
* 
Solution of Oblique Triangles 
Given 
Required 
b, c , O 
b = 
4* a, b 
B, c, C 
sin L 
a, K e 
a 
s — 
• 
* m 
1 
Area 
a 
b t 6 
8 = 
A % b . c 
area 
A, B, C, a 
Arch 
"* ^ , (7 =5 180 °. — (A + B),o = a s:n ° 
sin 4 
sin A 
— - — t C = 180°— {A + B) f c = -r — 7 
sin 4 
€ — 
d sin 6 V 
sin 4 
a -f d 
V * 
A d 
sin 
ac 
m°-(A + B) 
, area = \A(* — «J (s—c) 
sin 
a 1 sin B%\n G 
2 sin 4 "il "6 
reduction to horizontal ' * « 6 
Horizontal distance = Slope distance multiplied by the 
cosine of the vertical angle. Thus: slope distance -1119. ■« ft. 
Vert angle 5° 10 / . From Table, P^ge IX. cos 5° 10' — 
i> .9959. Horizontal distance— 319.4X. 9969 -31 8,09 ft 
r Horizontal distance also — Slope distance 1 minus slope 
same figures as in the preceding example, the follow- 
Horizontai distance mg result is obtained. Cosine 5° 10'—. 9959. 1— .9960 -.0041. 
319.4X.0041 = 1.31. 319.4 -1.31 -3ia09 ft, 
■ ' 
ance less the square of the rise divided by twice the slope distance. Thus: rise-- 4 it*, 
slope distance^ 302.6 ft Horizontal distance— 302.6 — i4 -- 302.6— 0-32 302. 28 ft. 
j 2_ 
MA0EIMU.S.A. 
