IG FORMULA 
Cl*- 
/ 
rianph 
For Angle 4. sin 
Given 
a, b 
Required 
4, B ,c 
Oblique Triangl 
lutlon of Right Triangles 
bah 
, cos's — tan = — C ot = 
lan A = — = cot B t c — y/ a* = 
sin A = — = cos B, b = yV (a-j-a) (<j — a) = , 
c 
B = 90° — A, b = acot4,c = — 
sin A . 
B y a, c 
t 
B, a, b 
B = 90 c 
, a = b tan A , c — 
Uiven 
Required 
cos A, 
= 90° — 4, a = c sin A, b = c cos 4/ 
Solution of Oblique Triangles 
rt ‘ iono /a i n\ «sin G 
V - 0 = 180 +. B )> « = 
sin 
^ - 
a 
C — 180° — (4 -f- B) t c = -r 
sin 
sin 4 
O. 
4 f Z? = 180° — (7, 
tan 
1 1 
a sin 
c — — — 
sin j, 
m (a— 6 ) tan j{A + B) 
a +. b 
A. h 
? /7 
sin ^4 
,siii 
j(s — b){s ■ 
b e 
r f ■ h 
A r 
a\ Ar 
K 
sin ?, 5 . O- lgO’-f A -t- 
a -b i 4 c 
= 9 . area — V »(*—«) — b) , 
6 c siq 4 
ea “ 2 
ea * 2 7 uT4 
CTION TO HORIZONTAL 
Horizontal distance =» Slope distance multiplied by the 
cosineofthe vertical angle. Thus: slope distance 319.4ft. 
Vert, angle — 5° 10'. From Table, Page IX. cos 5° 10'= 
.9950. Horizontal dis'tance=319.4X.9950— 318.09 ft. 
Horizontal distance ab>o- Slope distance minus slopp/ 
distance times (1— cosine of vertical angle). With >tfe 
same figures as in the preceding example, the*! 
itig result is obtained. Cosine f 
19.4X.0041 =1.31. 319.4 — 1.31 =31 8.054 ft 
lorizontal distance 
/hen the rise is known, the horizontal distance is approximately:— the slope dist- 
less the square of the rise divided by twice the slope distance. Thus: rise— 14 ft., ' 
ipe distance i=302.6 ft Horizontal distance— 302. f>^ =302.6—0.32^302.23 ft. 
■ »• \ 2 X 302.6 
. * ... MADE IN U, 8, A 
