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<br />TRIGONOMETRIC FORMULlE `
<br />B B = B
<br />c a ° a c a
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<br />{` RiAlit Triangle Oblique Triangles
<br />Solution of Right. Triangles
<br />For Angle A. sin 2'cos = b ; tin = 2 , cot = b ,sec = 6 , cosec = f
<br />e c b a b a
<br />4 Given Required a
<br />b A, B ,c fan d = -b = cot B, c = /as { b = m i +
<br />a, c A, B, b sin A= = cos B, h= c 1— o a
<br />A, a B, b, c B=90'—A, b = a cot A., c =
<br />sin A.
<br />A, b B, a, c B = 901—A., a = b tan A, e = .
<br />1 A, c "B; a, b B=9 ' 0*—A, a = c sin A, b = e s -A , -` X36 5
<br />Solutioof Oblique Triangles �� -
<br />Given,' .Required a sin 2sire
<br />B
<br />i A, B,a b;'. 6, C. ,b= C=180`—(A+B),0=C
<br />},sin A sin tl
<br />bsin d asi
<br />A, a b B, a, C'. sin B = a , C = 180 iA t B) , c = sin 4:
<br />a, b, C- A, B, c A+B=1801 C, tan "(A—B)= (u —b) tan 'z (:AFB)
<br />aSill C a+b
<br />sin A
<br />a, b, e A, B, C s= 2 sin A
<br />f sin:'-B=at'6 `),C=180'—(A+B)
<br />a, b, c Area s= 2
<br />44
<br />•A, b, o '.Aiea area = b c sin A�,-
<br />2 O e,
<br />=
<br />A, B, C,2 Area area = asin B sin C
<br />2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal disfance=Slope distance multiplied by the "� r
<br />ecosine of the verticalangle.Tbus:slopedistance=319.4ft. -
<br />tpre Vert. angle=51 101. From Table, Page IX. cos 51 10/= , (`
<br />e ass y 9959. horizontal distance=319.4X.9959=318.09 ft. �a
<br />$lop i,1c Horizontal distance also=Slope distance minus slope/: •.� -•
<br />distance times (1 -cosine of vertical angle). With the
<br />`i same fig -ares as in the preceding example, the follow- r0
<br />Horizontal distance ing result i"s obtained. Cosine 50101=.9959,l-.9959=.0041.
<br />319.4X.00,1=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=l4 ft.,
<br />slope distance=302.0 ft. liorizontal distance=302.6- 14 X l4, =3026-0.32=302.28 ft.
<br />2 X 202.6
<br />!AA9a IN U.B.A.
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