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<br />TRIGONOMETRIC FORMULAE
<br />B B B
<br />e a c a ° a
<br />b Q'AI b CA b �C
<br />tRight Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />a b a b c c
<br />'For Angle A. Bin = c , —=. o ,tan = b , cot = a , sec = b , cosec = —
<br />Given Required a
<br />a, -.b A;B,c tanA=b=cotB,e= a2+: a=a 1+ as
<br />a, c A, B, b sin A = a = cos B, b = o+a) (c—a) =0 J 1— o
<br />A, a B, b, c B=90°—A, b= a cotA, c= a
<br />sin A. ,
<br />A, b B, a, e B = 90°—A, a = b tan A, c = b
<br />cos A.
<br />A, a B, a, b B = 90°—A, a = c sin A, b = e cos A,
<br />r Solution of Oblique Triangles
<br />Given Required a sin B a sin
<br />_
<br />A, B, a b, c, C b sin A ' C = 180°—(A + B), — Bill A
<br />A,, a, b B, c, C sin B= b sa'A,C = 1800—(A + B), o = sin A
<br />a, b, C A, B, o A+B-1800— C, tan ? (A—B)— (a—b) tan ; (A+B)
<br />a+b
<br />a sin C
<br />c=
<br />sin A
<br />a+b+c —
<br />--A, B, C a— 2 ,sin�A—` be ,
<br />sin 2B_,�(8 a 0 C-1801--(A+B)
<br />g, b, a Area s= a+b+o2 ,ares = s(e—a s— s—c
<br />A, b, 'c Areaencs = b e sin.A
<br />2
<br />area = a9 sin B sin C
<br />A, 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 />roe Vert. angle =5° 1V. From Table, Page IX. cos 50 lot=
<br />e ass H 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />glop Angle ' Horizontal distance also=Slope distance minus slope
<br />Ve 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° 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=302.28 ft.
<br />2 X 3026
<br />wwe W U. & A4
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