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<br />TRIGONOMETRIC FORMULAE
<br />B B
<br />a a a
<br />b C b C C
<br />�Nht Triangl'e Oblique Triangles
<br />Solution of Right Triangles
<br />a b a b
<br />For Angle A. sin= ,Cos= tan =b a see cosec
<br />b a
<br />Given Required --- F2
<br />a,
<br />a, b A,jB,o tan b cot B, c = N/—a2
<br />1 + NF1 + 7a -
<br />Ai B, b sinA= o=cosAb= o2
<br />1'4!' 4' A b, 'e B=90°—A, b = a cotA, c= a
<br />sin A.
<br />A, b B, a, blan-A,-6 z�i
<br />B=90'—A,a--. '
<br />A.
<br />i, A, c B, a, b I B = 90*—A, a = c sin A-`-Y—�-.c-cos A
<br />Solution of Oblique. Tkaii-gAeris
<br />Given Required asin B "in,C
<br />A a b, c, C b =—
<br />sin A a=180*—(A + B)
<br />sin A'
<br />b sin Aa sin C
<br />�A, a, b B, b, C sinB= , C = 1800—(A + B),,
<br />b,.!g A, B, c A+B=180'— C, taiiNJ(A-
<br />a+ b
<br />b" . sin- C
<br />sin A'
<br />a+b+
<br />0 a, b, o A, B, C .,in 'A=�
<br />2 2 b c
<br />'k
<br />sinjB=Y(s—a. ac ) C=180*--(A+B)
<br />a+b
<br />'a, b, c Areas= area =s(8—a 8 (s—c
<br />besin A
<br />A, b,,c Area area = 2
<br />a2 sin B sin C
<br />B, C, a Area
<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 />ce, Vert. angle= 50 101. From Table, Page IX. cos 51 101=
<br />.9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />sva'oe S:oVXe .2 Horizontal distance also=Slope distance minus slope
<br />'t , 04 distance times (1—cosine of vertical angle). With the
<br />-4 el - 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=302.6— 14 X 14 =302.6-0.32=302.28 ft.
<br />2 X 302.6
<br />MADE a1 U. L 4
<br />-4
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