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TRIGONOMETRIC FORMULAE
<br />B B
<br />a C
<br />a a
<br />A A A
<br />f b C . b
<br />Right Triangle Oblique Triangles
<br />Solution, of Right Triangles
<br />For Angle A. sin=. cosCos= tan = a cot = b sec cosec
<br />a
<br />'Giyen Required
<br />2
<br />a,b A, B,e tanA cotB, c N/_a_2_+_7T2= a T
<br />a, oAt B, b sin A= a= cos B, b= \,/'?7c+ a) —(o—a) 0
<br />a
<br />A, a, B, b, e B=90°—A, b = a cot A, 0=
<br />-sin A.
<br />A, b B, a, e B=90*—A,a = btan A,c= -
<br />cos A.
<br />A, a B, a, b I B =90*—A, a = e sin A, b = e cos A,
<br />Solution of Oblique Triangles
<br />Given Required a sin B a sin C
<br />A, B, a b, e, C- b C =180'—(4 + B), e = —
<br />sin A ' sin A
<br />'B b� sin Aa sin C
<br />A, = C sin a , 0'= 180'—(A + B) , sin A
<br />a, b.:C A, B, -c A+B=180*— C, tan 1(4—B)= (a—b) tan (A+B)ab'
<br />2
<br />a sin C +
<br />sin A
<br />a, b, a A, B, C 8=a+b+ sin'A=
<br />2 be
<br />sin'2B= 8-_—)(s_),C=180---(A+B)
<br />N a 0
<br />a+b+c
<br />a, b, a Area S=v's�(,—a s-
<br />2 ,area =
<br />A, b, o Areaarea = b c sin A
<br />2
<br />trl �1� a2 sin B sin C
<br />A, A C, a Area area = -
<br />2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />'oee cosine of the vertical angle. Thus: slope distance =319.4ft.
<br />A� z Vert. angle =5' 101. From Table, Page IX. cos 50 101=
<br />..9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />Slop b:Horizontal distance also =Slor)e 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 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 />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=302.6- 14 X 14 ==6-o.32=302.28 ft.
<br />2 X 302.6
<br />MADE IN V. S. A.
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