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<br />TRIGONOMETRIC FORMUL./E
<br />B B - B
<br />c a c a c a
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<br />b -Ae Oblique C A b ` C
<br />'Right Triangle Triangles 1
<br />Solution of Right Triangles
<br />For Angle A. sin= a b a b c c
<br />g c , cos = 0 , tan = b , cot = a , sec = b , cosec =
<br />Given Required
<br />a,b A,B,c tanA=b=cotB,c= a= { a=a� 1� a
<br />z
<br />a,.c d, B, b sinA==co`sB,b=\I(c Fa)(c—a)=c1—o;
<br />A, a - B, b, c B=90°—A, b = a cotA, c= sin A.
<br />• .�1 . b
<br />A,' b B, a, c B = 90°—A, a = b tan A, c =
<br />cos A.
<br />A, c B, a, •b` B=900—A, a = e sin A, b= e cos A,
<br />a' Solution of Oblique Triangles
<br />Given Required a sin B — a sin C
<br />'A, B, a It, c, .0b' = sin A ' C = 180°—(A + B), c = sin A
<br />b sin d a sin C
<br />A, a, b B, c, C sin B= a ,C = 180°—(A (B), c = sin A
<br />�. a, b, .0 . A, B, c A+B=180 C, tans (A—B)= (a -b) tan Y (A +B)
<br />b
<br />a sin C a+
<br />c I= sin A
<br />a, b, c A, B, C s=a+2+c,sin 'A=�(s- a(c—c ,
<br />sin aB= \I C=180°—(A i B)
<br />' ac
<br />a+ b -i- c
<br />a, b, a Area S= 2
<br />A, b, c Area area bcsin A
<br />2
<br />A, B, C , a Area area = a2 sin B.sin C
<br />I' 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 />tar°o Vert. angle=51101. From Table, Page IX. cos 50101=
<br />e ais m 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />5
<br />No4e4p'�Ne t4 distan ettimest(lccosi a oflvertcal angle).1With lthe
<br />V same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5° 101=.9959,1 _.9959=.0041.
<br />f 1 319.4X.0041=1.31. 319.4-1.31=318.09 ft.
<br />hWhen the rise is known, the orizontal 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 />y_ 2 X 302.6
<br />MADE M U. 8. A.
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