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' TRIGONOMETRIC FORMUL/E
<br />B
<br />c a c a c a
<br />A b CA� b C,A C
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin = ,cos = ,tan = b ,cot = a ,sec = b, cosec = a
<br />Given Required a z
<br />a, b. A;B,o tan A=b= cot B,o= az z=a 1-}-dz
<br />z
<br />a, c A, B, b sin A = = cos B, b = \1-(c+ a) (c—a) = c 1— a a
<br />A, a B; b, c B=90°=A, b= a cotA, c= a
<br />sin A.
<br />A, b B, a, o- B =900—A, a = b tan A, c = b
<br />X 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
<br />A, B,a b, e, C b=asinB�C=180°—(A+B),c=asinC
<br />sin A sin A
<br />b sin A a sin C
<br />A; a, b B, o, C sin 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 />a + b
<br />c=
<br />a sin C
<br />sin A
<br />d, b, c A, B, C s=a+2+c,,inlA= Y(s— b
<br />,I
<br />sin'B= Y ac ,C=180°—(A+B)
<br />a+b+c
<br />i ca, b, ,c Area s= 2 , area = s(s—a) s— ) (s—c
<br />A, b, c Areaarea = b o sin A
<br />2
<br />a? sin B sin C
<br />�, 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=319.4ft.
<br />tQ,pe Vert. angle= 60 101. From Table, Page IX. cos 50 101=
<br />Pe ass m 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />o
<br />S,Ve�•br�,e a distance distance
<br />a oflverticalslope
<br />angle). With
<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 />1.
<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 '4-302.6-0.32=302.28 ft.
<br />2 X 302.6
<br />rwe IN U. $.A.
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