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?; 41 TRIGONOMETRIC FORMUL E—
<br />r"
<br />r� n a
<br />b
<br />Cb C b
<br />Right Triangle Oblique Triangles '
<br />Solution of Right Triangles'Fo
<br />o r Angle A. sin = a cos = , tan = ','cof = a , sec
<br />Given, Required a z a
<br />1 a, b A' B ,o tan A =.b = cot B, a a$ ''= a 1 + dz i
<br />eA, B, b ;sin A = =cos B, b = �J {e l a (e—a} = c 1— o
<br />Vi A, a B, b, c B=gQ°—A, b= atootA, o a 4
<br />-' sin A.
<br />A, b B, a, a B=90'—A, a= bran A, c b
<br />cos A
<br />rA,.c - rB, a, b= . B=90°—A, a;=.csinA, b= a coed,
<br />G Solution of Oblique Triangles
<br />{ i Given Required a sin B
<br />1 j A B,.a d, c, C b sin A C=.180°—(A + B), c sin A
<br />$b sin A , a sin C
<br />B, c, C sin B = a , t: = 180 -(A + B) , c sin A
<br />af, a, C A, B, c A+B-1$0°-0,tan;(�-8)= a—b)tan t1 B)
<br />1
<br />- E _ sine 4
<br />'J A, B, C 8-- . 2 in JA=_bo
<br />C=180°— (A+B)
<br />i.�
<br />a, b, . e Asea s = a+2+c, area = s{a ac a— (s c
<br />bosinA
<br />G., b, c Area area = .
<br />fi 2 ' S:
<br />�g B C a Area area — as sin B sin 0 a
<br />2 sin A ?
<br />REDUCTION TO HORIZONTAL ;I
<br />Horizontal distance= Slope distance multiplied by the . S
<br />e- cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />Igoe Vert angle= b° w. From Table, Page IX. cos b° lo'=
<br />S m 9959. Horizontal distance=319.4X.9859=318.09 ft i
<br />1 51o4e P4��e Horizontal distance also --Slope distance minus slope
<br />�e distance times (1 --cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow- i
<br />Horizontal distance ing resultis obtained. Cosine b° l0'=.9959.1—,9959= .0041.
<br />- 818.4X.11041—t.31. 518.4—].81=318.09 it
<br />Whenthe rise is known, the horizontal distanceis approximately.—the slope dist
<br />-
<br />�,nce less the square of the rise divided by twice the slope distance. Thus: rise=l4 ft,�,.
<br />dope distance=9029 ft Horizontal distance=3028— 4 X 14 =8026 932=30228 ft.
<br />2 X 9026 j
<br />MADE IN U.S.A.
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