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<br />TRIGONOM'16TRIC FORMULrE
<br />B B
<br />c c c
<br />a � a a
<br />A b CA�b CA' C
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin= a , cos = b , tan= a , cot =�b sec = o , cosec = e
<br />c c b a' b a
<br />Given Required
<br />a,b A,B,o tan A=a—b= cot B,c= a2+ z=a 1+ az
<br />a, o A, B, b sin A = � = cos B, b = \/ (c+a) (o—a) = c 1— 0
<br />A, a B, b, c B=90°—A, b= a cotA, c= a
<br />sin A.
<br />A, b B, a, c B = 90°—A, a = b tan A, c = b
<br />.cos A.
<br />A, c B, a, b I B = 900—A, a = c sin -A, b = c cos A,
<br />Solution of Oblique Triangles
<br />Given Required a sin B
<br />A' B,a b, c, C b= C=180°—(A+B),c=asinC
<br />sin A ' sin A
<br />A, a, b B, c, C sin B = b sin A, 180°—(A +B) , c = a sin C ,
<br />a sin A
<br />a, b, C A, B, c A+B=1800— C, tan -,X (A—B)— a—b) tan s (A+B)
<br />—
<br />c=
<br />a sin C
<br />sin A
<br />A, B, C s=a+2+c,sin;A=
<br />sin zB=�(s—aa(a ),C=180°—(A-i-B)
<br />a, b, c Area s=a+2+0, area = s(s—a (s—)(s—c
<br />A, b, c Areab c sin A
<br />area = 2'
<br />A, B, C, a Area area = as sin B sin C
<br />2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied by the
<br />e cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />a ita c Vert. angle=50
<br />101. From Table, Page IX. cos 60 lry=
<br />9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />S&I a Arg1e Horizontal distance also =Slope distance minus slope
<br />Ve distance times same figures (1—cosine of vertical angle). With the
<br />gures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 60 101=.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 =302.6-0.32=302.28 ft.
<br />2 X 302.6
<br />MADE IN U. B. A.
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