D . i . <* if a o c c 
r or Angie A. sin = - , cos — — f tan = — , cot = “ , sec = - cosec *= — 
a i 
Given 
a> c 
A, a 
A, b 
A, c 
Given 
A, B , a 
A, a, b 
a, b 9 C 
Required 
A , B ,d 
A. B. b 
B, b, e 
9 r* 1 ■ 
tan il = 7 - = cot B, c = y a ~f /T ** <* -y 1 *+* 7 
■ 1 | * 
sin A = — «* co 8 j B>*b = \/(c + a)(c — a) = c •> /1 — — 
d V ‘ 
/? = 90 — id, b = a cot A f d = 
sin A. 
B } a, c —A 4 tan A , 
w 
B y Oy b B = 90°— A, a = esin A, b=- c cos A, 
Solution of Oblique Triangles 
a, b f 0 
<k b, e 
A . b. c 
A, By C, IS 
Required 
A, d, <7 
i?, d, £ 
*4, B . d 
A^ By C 
•1 
Ares 
Area 
Area 
<1 sin C' 
£4 sin 
* - ~t r»^ =* 180 —(a + ^>, * - 
Bin -4 am A 
sin B = - f (7 = 180 — (A-f- Zri, d =-— 
in A 
^t+^=180°— C, tan !(/—£) = l a —'l tan L<A±g) 
a.inC “ + * 
d = 
<H-*+c , 
HT1 
sin A 
l*——g) 
be 
,sin \A=^ 
\ a e 
= ^iA±f # area = \' r *{*- A) (j— }>) (.»— c > 
area = 
area = 
b c sin A 
a 3 sin 2? sin (7 
Horizontal distance 
2 tin 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^ 
<u .9959. Horizontal distance 319.4/ 9950 3lK D9fL 
Horizontal distance also Slope distance milium slot*© 
distance times (l—co 3 ine of vertical angle). With the 
sams figures is in the preceding example, the follow¬ 
ing result is obtained. Cosine 5° 10 3 — 9969.1—.9959=.0041. 
319.4 X 0041=1.31. 319.4-1.31=318.09 ft. 
WheQ the rise is known, the horizontal distance) - approximately: the dlst* 
ance less the square of the rise divided b y t wice the slop 1 distance. Thus: nse=14 ft., 
slope distance=3*12.6 ft Horizontardlstance—302.C—=302,0—0.3?-.'W2.28 ft 
2 302.6 
■W 
14ADC m U O At 
