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3 7 7 8 °` _ A, B, C, a Area area = as 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 />e a�sti9�co d .99959. Holr Horizontal distance 3Table.
<br />9 4X 9 9e 315.09 ft.IX. cos 5° ilY
<br />e,1op Angle 4 Horizontal distance also=Slone distance minus slope
<br />a distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow-
<br />ing result is obtained. Cosine 5° 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=3026 ft. Horizontal distance=3026— 14 X 14 ==6-0.32=30129%
<br />2X3026
<br />IME in U. a.
<br />TRIGONOMETRIC FORMULYR
<br />B B
<br />b�i
<br />z
<br />a c a °. a
<br />9i1 i'."334.2/
<br />A
<br />I,r
<br />/ Z '. - _
<br />_,
<br />f
<br />Right Triangle Oblique Triangles
<br />W
<br />i _ j •� �/(�
<br />l� G
<br />-,., Solution of Right Triangles
<br />a b a b a a
<br />!
<br />_
<br />tan= T, cot
<br />For Angle A. sin = coe = ,cot = ,sec = b , cosec =
<br />�t
<br />/
<br />7'��O
<br />_ � -
<br />Given
<br />a,b
<br />Required
<br />A,B,a
<br />e-, a a a
<br />a a
<br />tanA=b=cotB,c= a2+.'=a 1+aq
<br />3,
<br />a
<br />A, B, b
<br />sin A = eos B, b = c a c—a = c I—;—,
<br />f
<br />A, a
<br />B, b, c
<br />a
<br />B=90°—A, b = a cotA, c= sin A.
<br />�o
<br />7-
<br />A, b
<br />B, a, c
<br />B=90* —A, a . = b tan A,
<br />/
<br />cos A.
<br />D
<br />'2
<br />A, c
<br />B, a, b
<br />B = 90°—A, a = c sin A, b = a cos A,
<br />Solution of Oblique Triangles
<br />Given
<br />A B,a
<br />Required
<br />b c, C
<br />a sin B a sin C
<br />b= ,C=180°—(A+B),c=
<br />sin A sin A
<br />A, a; b l
<br />B, c, C
<br />b sin A ° a sin C
<br />sin B = , C = 180 —(A + B) , c =
<br />A
<br />a, b,sin
<br />C
<br />d., B, a
<br />A+B=180°— C, tan ; (A—B)= (a—b) tan a (A+B)
<br />` -
<br />6{zo
<br />12
<br />jjj
<br />r
<br />-
<br />f
<br />a + b
<br />_asin C
<br />c sin A
<br />f,
<br />-
<br />!
<br />k
<br />0. b, a
<br />A, B, C
<br />a= '+b+" 'A
<br />b(c—c
<br />/0
<br />/oqs
<br />J
<br />sin4,B=`Is`ac ,C=180°—(A+B)
<br />38-a
<br />b, a
<br />Area
<br />s _ 2 ,area = a(s—a a— s—c -
<br />ZZ / 94
<br />A, b, c
<br />Area
<br />bosinA
<br />area =
<br />2
<br />3 7 7 8 °` _ A, B, C, a Area area = as 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 />e a�sti9�co d .99959. Holr Horizontal distance 3Table.
<br />9 4X 9 9e 315.09 ft.IX. cos 5° ilY
<br />e,1op Angle 4 Horizontal distance also=Slone distance minus slope
<br />a distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow-
<br />ing result is obtained. Cosine 5° 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=3026 ft. Horizontal distance=3026— 14 X 14 ==6-0.32=30129%
<br />2X3026
<br />IME in U. a.
<br />
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