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<br />TRIGONOMETRIC FORMUL)E
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<br />Right Triangle Oblique Triangles
<br />( t Solution of Right Triangles
<br />Angle A. sin=
<br />a b a
<br />For An l b c e
<br />g cot = a , sec_ = b , cosec = —
<br />I Given Required 7S %a
<br />a,b A,B,c land=b=cotB,c= az+ 2=a 1+ s
<br />�I� a
<br />t li, a, c d B, 'b sin A = = cos B, b = \/ (c+a) (c—a) = c V 1— o
<br />t A. a B, b, c B=90°—A, b= a cotA, c= a
<br />sin A.
<br />' A, b B, a, c B = 900—A, a = b tan A, c = cos A.
<br />A., c B, a, b I B=90'—A, a = e, sin A, b = c cos A,
<br />Solution of Oblique Triangles `
<br />j Given Required a sin B
<br />!d, B, a b, c, C b sin d ' C = 180°—(A + B), c = sin A
<br />b'sin d a sin C
<br />'A, a, b B, .c, C in B = a , C = 180°—(A + B), c = sin A
<br />a, b, C A,' B, c A+B=180°— C, tan (A—B)= (a—b) tan (A+B)
<br />'ab '
<br />a sin C +
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<br />sin A
<br />a, b, c .A, B, Cs:=a+b+c,,in;A= Y(s a(c
<br />sin i B= a(•C c) C=180°—(A+B)
<br />a, b, o 'Area s = a+2 +c, area
<br />• t;A, b, e, Area b c sin A
<br />area = 2'
<br />a2 sin B sin C
<br />A, B, C, a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied by'the
<br />cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />oGe ass«ree Vert. angle= 50 101. From Table, Page IX. cos 50 Hy
<br />9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />51 N41 94 a Horizontal distance also = Slope distance minus slope
<br />ire t distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5° 101=.9959.1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=318.09 ft. '
<br />When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise=14 ft.,
<br />slope distance=302.6 ft. Horizontal distance=3026— 14 X 14 =30.6-0.32=302.28 ft.
<br />2 X 302.6
<br />AWE IN U.S. a
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