1 
TRIGONOMETRIC FORMULAE* * 
B 
n 
B 
c 
Right 1 Vi angle 
b G ' b 
Oblique Triangles 
G 
Solution of Right Triangles 
a b a b 
For Angle A. sin = - , co«= - , tan = - , cot = - , sec = f , cow - -- 
Given 
* M 
Required 
A, B ,o 
*> 
A, B b 
A, a 
B , b , c 
A, b , 
B % a % e 
A. c 
B y a. b 
. Given 
Required 
tan A = -- = cot i,s = v' 0 ’ r = a l + y 
a, b, a 
sin A — — = cos B , b = \/ (c-\-a) ( c — a) =* c * / 1 — ~ 
A 
5 - 90 ° — A t b « a cot A , c « 
/? — 90 ° — A , a, ~ b tan , c = — — 
cos ^4/ 
B *= 90 u — .A, a = c A, b =- c cos A t 
b = t c = 180° — (A + 5). c «- 
sin A 
♦ 
sin 2? = — in — - % G = 180°— (J. -p 2?), c = 
sin v4 
a sin C 
sin *4 
A % B> G 
Area 
.4+5=180°— C.tan UA-£) A a ~ b ) t * n i<4±g) 
' __ a tin C a + b 
• sin A 
g+bj-c /OKfED 
2 • J "V ie 
sin <7=180°— M -t-5) 
, area = V«(a — <*) (* — b) (s— c) 
area — 
a’ sin B sin (7 
e®. 
2 sin A 
REDUCTION TO HORIZONTAL 
Horizontal distance = Slope distance multiplied by the 
cosine of the vertical angle. Thus: slope distance 319. 4 ft. 
Vert angle — 5° 10'. From Table. Page IX. cos 5° 10' = 
« .9959. Horizontal distance- 319.4X 9950=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- 
* 319.4 x .0041 = 1 4—1.31 -318.09 ft. 
When the. rise is known, the horizontal distance is approximately:— the slope dist» 
ttnoe Hie square of the rise divided by twice the slope distance. Thus: rise 1 i ft, 
slope distance— 302.6 ft. Horizontal distance^-302.6— ^ 4 - >Q! ^302.6— 0 32= 302.28 ft. 
2 X 302.6 
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MADE IN U. 8- h. 
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