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TRIGONOMETRIC FORMULlE
<br />B B
<br />C a Ck a C a
<br />b Cdl b CA b C
<br />Right Triangle Obli't)ue Triangles
<br />Solution of Right 'Triangles
<br />a b a b c
<br />For Angle A. sin= b 'cot= a, sec = , cosec =
<br />a
<br />Given Required a a
<br />a, b A, B ,c tan 1. = b = cot B, c as -1-
<br />a
<br />a, e A, B, b sin A = = cos B, b = �/ (c=a) (c—a) ay
<br />=c 1—a
<br />A, a B, b, r,
<br />sin A.
<br />A, b B, a; 'e B = 90--A, a = b tan A, c = b
<br />cos A.
<br />A, c B, a, b B=90°—A, a = 0sin _l, b= e.cosA,
<br />Solution of Oblique Triangles
<br />Given Required a sin B
<br />A, B; a b, c, C b = sin A ' C = 180°—(A + B), e = 'Sin C
<br />sin A
<br />A, a, b B, c, G' sing= b sin A ,C=180'—(fl.-t-B),a sin C
<br />a c= sin
<br />a, b, C A, B, c A+B=180°— C, tan 2 (A—D)=. (a—b) tan si (A+B)r
<br />a sin C a + b
<br />c=
<br />sin A
<br />A, B, G' s = —2,sin 1,A=-1 b o
<br />� (y—a)is—cl
<br />sin oB= ,C=130'—(A+B) .
<br />ac
<br />ac
<br />a, b, a Area a+b-I= 2 ,area
<br />A, b, c Areab c sin A
<br />urea = 2
<br />a' sin B sin C
<br />A, B, C, a Area area = 2 sin t,
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />e cosineofthe verticalangle.Thus: slope distance =310.4 ft.
<br />^e Vert. angle=5° 10'. From 'fable, Page IX. cos 61101=
<br />VC a�stia M 9959. Horizontal distance=3l9.4X.0950=318.09 ft.
<br />$lotogle ix Horizontal distance also=Slcpe distance minus slope
<br />distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5"10—.9959.1—.9959=.0041.
<br />319.4X.0011=1.31. 310.4-1.31=313.09 ft.
<br />When the rise is known, the horizontal distance is approximately:—ttie slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise=l4 ft.,
<br />slope distance=302.6 ft. Horizontal distance=302.6— 14 X 14 =303.6-0.32=302.29 ft.
<br />2 X 302.6
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