TRIGONOMETRIC FORMUL>E 
/; 
77 
_ a b a b 
For An^ie A. sin = — , coa — - , tan = t , cot = ~ , see 
C C 0 d 
Given 
a. b 
•jf 
Req lured 
A, B .r 
tan A =i — — col B, b — \/a* -j- ^ 
sin A = ^ = cos B^b = v/(c-|-d) (< 3 — a) =» 0 ^ 1 — ~ 
^ = 90° — A, b = acot-A,c— — 
B = 90 ** — A , a = ft tan , c = — - 
cot A. 
B = 90 ” — A — c sin ^ , ft = c cos A , 
Solution of Obliquo Triangles 
a 
o, ft. 0 
1' eqvured 
ft. € 
A, B, a 
b, c 
, 
Areg 
a = - 
u-i. ft, c 
Area 
area 
A, B, 0, ti 
Area 
area 
b = (7 = 180°— M + B), e = 
sin A 
ft sin A 
sin 
B^ 
,0= 180”— ■i^B),o = 
dsin C 
sin A 
a sin O 
sin A 
A+B=m°- C, un l(A-n)= ^“ 
asinC ^ + ‘ 
C = r 
Bin 
a-fft + c . . ^ ftH«— c) 
= -^— .»mi.4 = y 
sin i , <7= 180°— < A.^if . 
a c 
, area = y/ s (« — o) ( J — ft ) ( 
ft 6 sin A 
2 
o* tin 77 gin O 
2 sin ^ 
REDUCTION TO HORIZONTAL 
Horizontal distance — Slo(*e distance miillipHcd by lha 
cosine of the vertical an Hrle. Thus: slope distance ^31U. 4 ft. 
Vert, angle ^6^* 10', From Table, Page IX. cos 6^^ 10^ = 
« .9950. Horizontal distance -319.4X.9B59=318.09 ft. 
2 Horizontal distance al.so — Slope distance minus slope 
^ distance times (1— cosine of vertical angle). With the 
same fiirnre.s as in the preceding example, the follow- 
Horizontal diKtanMi result is obtained. Cosine S'* 10'-. 9969. 1— .9959— .0041. 
exunzonuu aisiancc 319.4X.0041 -1.31. 319.4- 1.31 -318.09 ft. 
Wbeii the rise Is known, the horizontal distance is approximately:— the slope *itsl- 
siice less the square of the rise divided by twice the slope distance. Thus: rise - U ft, 
' slope 'distance=902.6 ft. Horizontal distance^302.0-r - ILK I*. =^302.fi— 0.32^3(rJ-28 ft. 
2X302-6 
f 
i 
MAPE in U. S' A. 
