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TRIGONOMETRIC FORMULfE
<br />B B
<br />a ° a c a
<br />.,� b C A�b C A 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 = a
<br />c c b a b a
<br />Given Required a %' a
<br />a,b
<br />A, B'0 tanA=b=cotB,c= a2+ 2=ca 1-f ,
<br />a, o A, B, b sin A = u = cos B, b = c a sa
<br />c �( -I- )(c—a )=o�l—os
<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 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, c, C b = sin A , C = 180°—(A + B), c = sin A
<br />b. sin A a sin C
<br />A, a, b B, c, 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)=
<br />a+
<br />cb
<br />=
<br />a sin C
<br />sin A
<br />a, b, c A, B, C s=a+2+c,'iniA= Y(s _a(G
<br />sin;B=aa(c c ,C=180°—(A+B)
<br />a+b+c
<br />a, b, c Area 8= 2 , area
<br />A, b, c Area b c cin A
<br />area = 2'
<br />A, B, C, a Area area = W sin in sin C
<br />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 />Sta�'0e Vert. angle =5° 1V. From Table, Page IX. cos 50 i(Y=
<br />e a� y .9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />slop Angle Horizontal distance also=Slope distance minus slope
<br />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 50 10f=.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 M.
<br />slope distance=302.6 ft. Horizontal distance=302.6— 14 X 14 =802.6-0 32=302.28 ft.
<br />2X 302.6
<br />MADE IN U. B. A.
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