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TRIGONOMETRIC FORM'UL1R
<br />B B ;4
<br />4 A A b C
<br />I. Right Triangle C Oblique Triangles
<br />Solution of Right Triangles
<br />a b a b c
<br />I or Angle A. sin = o , cos= o , tan= , cot = a , sec = a , cosec =
<br />Given Required a
<br />a,b A, B,c tan A.—b=cot B,c= a.+b==a y�- QZ
<br />1
<br />a, c A, B, b sin A
<br />A, a. B, b, c B=90°—A, b= a cotA, c= a
<br />sin A.
<br />-
<br />A, b B, a, o B = 90°—A, a = b tan A, e = h cos A.
<br />A, c B, a, b I B = 90°—A, a = c sin A, b = e cos A,
<br />Solution of Oblique Triangles
<br />Given Required
<br />A B a -b, c, C b= a sin B C— 180°—(A +B), 0= a sin C
<br />sin A sin A
<br />A, a, b B, c, C sin B= a b sin A ,C = 180°—(A -¢ B), c = a sin C
<br />sin A
<br />(a—b)
<br />a, b, C �A, B, 0 A-�B=180 — C, tan z (A—B)= ten a (A+ B)
<br />+ h ,
<br />a sin C
<br />C =
<br />sin A
<br />a, b, c A, B, C s=a+,sin A—
<br />sin'zB= a(�
<br />•,'a, b, a .Area s=a+&+0, area =•\/s(s—a) (s— )(s—c
<br />A, b, e Area area = 8 c sin A
<br />2
<br />A, B, C, a Area area = az sin B 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 />oVe a�5ti$tlpe w Vert. angle=51 M From Table, Page JX. cos 50 10'=
<br />9959. Horizontal distance=319.4X.9959=31&09 ft.,
<br />til Kra a. Horizontal distance also=Slope distance minus slope
<br />We distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow.
<br />Horizontal distance inuresult is obtained. Cosine 5°101=.9959,1—,0959=A041.
<br />319.4X.6041=1.31. 319.4-1.31=31&99 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: risc�14 it.,
<br />slope distance=362.6 ft. Horizontal distance==G— 14 X 14 =302.6-0 32=302 28 ft
<br />2 X 902.6
<br />_ IdI�E IN U. BSA,
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