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<br />TRIGONOMETRIC FORMUUE
<br />B
<br />a a a
<br />�4 A . b C A b C
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />aw
<br />`For Angle A. sin:' ,cos = tan= cota see cosec
<br />Given Required. 2
<br />b
<br />'A B,C tan Ab cot B, c a
<br />+
<br />az
<br />q, c '-4B, b sin A =- a =cosB,b=V(c+a)(c._—_a)=C —a,—
<br />OS
<br />A, a B, b, c B=90 --A, b= acotA, c= asin A.
<br />b
<br />A b B, a, c B=90'—A, a = b tan A, c
<br />c.. A.
<br />A, a B, a, b I B=90'—A, a = c sin A, b= e cos A,
<br />Solution of Oblique Triangles
<br />Given Required- asinBasin U
<br />A, B, a 0 b = , C = 180°—(A +B), C =
<br />sin A sin A
<br />-sin B = , b sin AC 180'—(A + B), c = a sin
<br />A, a, b B;1c, a sin A
<br />(a—b) tan !'j (A+B)
<br />b, C -A, B, C A+B=180*—C, tan !,(A—B)= d + b
<br />a sin C
<br />sin A
<br />+b + b)8)
<br />a, b, a A, B, C 8=a 2 ,in -2A=�(8— —0
<br />b(c
<br />sin 1B— a o C=1800—(A+B)
<br />b, c Area s— 2 a -'-b+ , area'= (8— �J(8_c
<br />b o sin A
<br />A, b, el Area 2
<br />as sin B sin C
<br />B, C, a Area, area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />'ce cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />11), "'r, Vert. angle = 51 101. From Table, Page IX. cos 50 I(V=
<br />elt a) M59. Horizontal d!stance=3l13.4X.!)959=3l8.Mft.
<br />SoVe
<br />all .�2 diHorizontal distance also=Slove distance minus �ope
<br />stance times (1—cosine of vertical angle). Withlthe
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine -51 101=.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 />anci less the square of the rise divided by twice the slope distance. Thus: rise=14 ft.,
<br />slope distance=302-6 ft. Horizontal distance=302.6— 14 X 14 =3016-0.32=802.28 M
<br />2 X 302.6
<br />JL-
<br />MADE IN U. LA.
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