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IM M
<br />(TRIGONOMETRIC FORMULtE
<br />a e a
<br />b A�rb c A b }
<br />Right Triangle Oblique Triangles ——J
<br />Solution of Right Triangles
<br />a b a b c a
<br />For Angle A. si = i o` , cos — o , tan = b , cot = a , sec = b , cosec =
<br />a
<br />Given Required
<br />a, b A, B ,c tan A — b = cot B, e = a$ + x = a 1 } as
<br />a, a A, B, b, Bin A = = cos B, b = c a c—a ¢r
<br />C t%(=a i-- a,
<br />A, a B, b, c� 1 B=90°—A, b = a cotA, c= sin A.
<br />fw b
<br />A, b B, a, c B =90P—A, a = b tan A, c — cos A.
<br />A, c B, a,' b B = 90°—A, a = c sin A, b= e cos A,
<br />Solution of Oblique Triangles
<br />Given Required a sin B a ein O
<br />A, B, a b, A ' C 180°—(A + B). e = sin A
<br />A. a, b B, c, 1 C' Bio B = a b Bin A, C — 180°—(A f B), c = a ein C
<br />sin A
<br />I;
<br />a, b, C A, B, c A+B=180°— C, tan } (A—B)= (a—b) tan
<br />+ (A+B)
<br />a sin C
<br />7' o Bin A
<br />a, b, a A, B,iC s=a+2+aysin j — FE -77(l —
<br />Ias+b+e
<br />a, b, c Area 8— 2 , area
<br />A, b, a Area area = b o Bin A
<br />2
<br />A, B, C, a Area area = at sin B sin C
<br />2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied bythe
<br />cosine of the vertical angle. Thus: slope distance =918.4 ft.
<br />woe Vert. angle= 5a ld. From Table, Page IX. cos 51 10/=
<br />m 9959. Horizontal distance=919.4X.9959=918.09 ft.
<br />�aqW Horizontal distance also= Slope distance minus slope
<br />ope ��'
<br />distance times (1—cosine of vertical angle). With the
<br />1, Ve same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 51 101=.9959. 1—.9959=.004L
<br />9114X.0041=1.91. 319.4-1.31=31909 ft.
<br />When the rise Is known, the horizontal distanee is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise=l4 ft.,
<br />slope distance=302.8 ft. Horizontal distance=3026--14X14=3M".32=3M28 ft.
<br />2X9028
<br />1
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