**> N " , 
TRIGONOMETRIC FORMUL/E 
For Angle A. sin = 
, cos = — , tan= cot = — , sec = - 
c o a 
Given 
Required 
a, b 
A. B, e 
a, c 
A, B, b 
A, a 
B, b, c 
A, b 
B, a. c 
A, c 
B. a. b 
Given 
Required 
A, B, a 
b, c, C 
( 
, • A, a, b 
*»“i % . s 
B, c, C 
a, b, C 
A, B, c 
a, b, c 
A, B, C 
b, c 
Area 
A, b, c 
Area 
A, B, C, a 
Area 
tan A =-f- = COlfl, c= V a 2 + b 2 = a \J 1 + 
sin A = _ ^ _ = cos B, b = \(c + a){c-a)=c i a 2 ' 
B - 90° - A, b = a cot A, c = 
B = 90° - A, a = 6 tan A, c = — ^-r- 
cos A. 
B = 90° - A, a = c sin A, 6 = e cos A, 
Solution of Oblique Triangles 
, a sin B „ „„„ , , a sin C 
sin A > C - 180 - (A + B), c - S j n ^ 
S inB = AsinA. c=180 o_ (A + B) _osin^ 
a ' sin A 
A + B = 1 80° - C, tan Vi (A-B) = (o ~ h) <an Vi (A + B) 
a + o 
_ a sin C 
c _ sin A 
sin Vi A = 
sin V4 B = 4 / (»- attend . C = 180°.- (A + B) 
V a c 
V s(s - a)(s - i)(s -c) 
a + 6 + c 
2 > area 8 
b c sin A 
a- 2 
■ a 2 sin B sin C 
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 = 
J> 319.4 x .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 following result is obtained Cosine 5° 
10=9959 1 - 9959= 0041. 319.4x .0041 =1.31. 3T9.4 - 
1.31 =318.09 ft, 
When the rise is known, the horizontal distance is approximately: - the slope distance 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 - ■ 1 4 - j ~ = 302.6 - 0.32 - 302.28 ft. 
2 x 402,0 
Horizontal distance 
MADE IN U S. A 
