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<br />TRIGONOMETRIC FORMULAE
<br />B B B
<br />e a c a e
<br />a
<br />A b C Ab C .A b C
<br />Right Triangle Oblique Triangles
<br />( Solution of Right Triangles
<br />M a b a b c e
<br />For Angle A. ein tan = b ,cot = 2 ,sec = b , cosec = —
<br />a
<br />Given Required
<br />•ilA a; b A, B ,c tan A = — cot B, c = az 2 = a 1
<br />i.' -1- a2
<br />a, a A, B, b sin A=c =cos B,b=\/(c-1-a)(c—a) =c �1-0�
<br />a B, b, •c B=90'—A, b = a cotA, a= s� A
<br />J
<br />A, b B, a, e. • B = 90°—A, a = b tan A, c = b
<br />coi A.
<br />y A, c B, a, b I B=900—A, a = c sin A, b= c cos A,
<br />n`
<br />Solution of Oblique Triangles
<br />Given Required _ a sin B -sin C
<br />js A, B, a b, c, C b sin A ' C = 1800—(A + B), c = sin A
<br />A,a,b B,c,C sin B=bsin aA,C=180°—(A+B),c= sin
<br />�.
<br />a, b,AC A, B, c A+B-180°— C, tan I (A -B)— (a—b) tan s (A+B)�
<br />e=asin C a+
<br />sin A
<br />a, b, a A, B, C s—a+2+c,Sin jA=� . b(G—c
<br />I,
<br />sin�I
<br />- ;B= Y ` C=180°—(A+B)
<br />' 2C
<br />a b+e
<br />a, b` Area s— +2
<br />` A, b, c Are- area = be sin A
<br />2
<br />i, A, B, C, a Area I area = a2 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 />s«npe Vert. angle= 50 101. From Table, Page IX. cos 50 101=
<br />H 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />ca�oPe AnQ1e z Horizontal
<br />ticoflerclan)1tltetnedistance
<br />es(cne ovtiagleWihh
<br />ale same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 50 101=.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 ft.,
<br />slope distance=302.6 ft. Horizontal distance=3026— 14 X 14 =3026-0.32=302.28 ft.
<br />` 2X3026
<br />MADE IN U. 9. A.
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