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TRIGONOMETRIC FORMULIE
<br />73 B B
<br />c a ° a c a
<br />d b C A b C A b C
<br />Right Triangle , Oblique Triangles
<br />Solution of Right Triangles
<br />a b }- b c c
<br />' For Angle A. sin = c , cos = , tan = a b , cot = ¢ —,see= b , cosec = a
<br />Given Required
<br />72-
<br />a,b A,B,c tanA=b=cotB,c= a2+ 2=a 1+a2
<br />a, c A, B, b sin A = = cos B, b = V/ (o+a) (c—a) = c 1— _;2
<br />4,.a B, b, a B=90°—A, b= acotA, c= a
<br />sin A.
<br />! A, b B, a, c B = 900—A, a = b tan A, o b
<br />cos A.
<br />}A,.c B, a, b I B = 90°—A, a = c sin A, b = c cos A,
<br />Solution of Oblique Triangles'
<br />Given Required a sin B a sin C
<br />A, B, a b, e, C b = sin A ' C = 180°—(A + B), c = sin A
<br />b sin A a sin C
<br />a, b B, c, C sin B= a , C = 180°—(A + B), c = sin A
<br />a, b, C A, B, e ,-A+B=180°— C, tan 1(.4—B)=(a—b) tan z (A+B)�
<br />a b
<br />a sin C +
<br />'r o=
<br />sin A
<br />a+b+c
<br />a,. b, 'c A, B, C s= 2 ,sin;A= be '
<br />I
<br />sin 2B= a c ,C=180°—(A+B)
<br />a+b+c
<br />a, b, c Area S= 2
<br />A, b, c Area..b e sin A
<br />area = 2
<br />- A, B, C, a Area area = a2 sin Bain C
<br />2 sin A
<br />REDUCTION TO IiORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />cosine of the vertical angle. Thus: slope distance=319.4ft.
<br />�S�aope Vert. angle =5° 101. From Table, Page IX. cos 5° I(Y=
<br />Pc a H 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />Sto Anse a Horizontal
<br />ttit(colercminusistance slope
<br />meslesine ofvtalangle).Withte
<br />�e same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5°10'=.9959.1—.9959=.0041.
<br />i 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 =3026-0.32=302.28 ft.
<br />2 X 302.6
<br />• xAoe u+ U. S.A.
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