TRIGONOMETRIC FORMULAE 
* 
' Given 
A, B x a 
A , #, b 
ck b, 0 
a, b, «, 
Reqyired 
K e, C 
B,e, C 
<4. B, e 
A, B , C 
a, b, e 
Area 
' A , b, c 
Area 
t 
A,B,C y a 
Aren 
b = a -S!S_^ _ G = 180° — (A + B), e = ^ ,n ° 
sin A 
•in A 
n f> sin A „ , a . rr. a 8 > n C 
sin B = t U =5 180 — ( A -f- H) . c = — — 
a sin A 
A+B=l&0°-C,lan \ iA—B) A a ~ b ) lan ^4+^? 
« (in C • + * 
(in A 
_q±±±e , Us-bXj-e) 
2 ' 1 "V be 
sin , 0-180°— I A + B ) 
= * : , area = \/*(» — O) (s— i) (*^e) 
• = 
area 
£ <J sin -4 
a* sin B sin O 
area — 0 . A 
l 6in A 
REDUCTION TO HORIZONTAL 
Horizontal distance = Slope distance rnultlplied by the 
cosine of the vertical angle. Thus: slope distance =319. 4 ft. 
Vert, angle — 5° 10'. From Table, Page IX. cos 6° ICK= 
£ .9959. Horizontal distance^319.4X.9959=31&09 ft. 
r 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- 
ing result is obtained. Cosine 5° 10'-*. 9959.-1— .9963=. 0041. 
319. 4 X. 0041 =1.31. 319.-!— 1.31 =318.09 ft 
o-M ' '' f i * \ r. ir ? ■ : r. ( r, 1 f i • r .>i« adores iij-ir.tely: rite slop-: dist- 
■Tr • . nco ‘I i :s. ~ 1 - 1 ; 
slope distance=302.6 ft Horizontal d istance -302.6- -302. 0. 32 - 302. 28 ft 
< 4 r 2X302 6 _______ 
Horizontal distance 
