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TRIGONOMETRIC FORMUL&
<br />gid? A e a
<br />bb G
<br />Right Triangle C L-7_ Oblique Triangles
<br />Solution of Right Triangles.
<br />'For An le'A, sin = a b ° t
<br />g C , coe = c tan= b , .cot = _a ,sec = b , cosec = a
<br />I (liven Required
<br />!F • ;,, a, b. , meq, B ,e tan A _ � = cot B; c ='V -5a +- 1+
<br />i
<br />a, e
<br />A, a _B, ii, c B=90°=A, b = a cotA, c= ¢
<br />' sin A.
<br />b
<br />"A, b • B, a, c B = 90°-A, a = b tan A, e = coe A '
<br />A, c B, a, ',b B = 94°—A, a = c sin A, b =, c cos A,
<br />Solution' of Oblique Trianglea
<br />t i Given Required a sin B ¢sin C
<br />A, B, a b, c, C" b = sin A ' C =,1.80°—( sin A
<br />b sin A a sin C. .
<br />A, a, b B, c, C . sin B = . a , C = 1800—(A + B) , c = sin A
<br />a, bC A, B, c A+B=180°— C, tan (A—B)= (¢—b) tan ; (A+.B) +
<br />ab
<br />� _¢ sin C
<br />�l ein A
<br />I a, b, a A, B, C s — ¢+�+e,sin jA=�(y
<br />sin3B= ac ,C-180°—(A+B)
<br />a, b, c Areaa — a+�(3— ) (.v—c)
<br />A, b, c Areab c sin A
<br />area — 2
<br />�A, B, t ; a Area arra =
<br />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 iv5��ce "Vert. angle=ti° 10+. From Table. Page IX. cos 51 I(Y=
<br />8959. Horizontal dislance=319.4X.9959=919.09 ft.
<br />Slop A�Ile iyX Horizontal distance also=$lope distance minus slope
<br />ye 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 50 10'=.9959.1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=318.09 it.
<br />j When the. rise is known, the horizontal distance is approximately:—the slope dist-
<br />ince less the square of the rise divided by twice the slope distance. Thus: rise=14 ft,
<br />" slope distance=302.8 ft Horizontal distance=02.
<br />3e— 14 X 14 =3028-0 32-302.28 it.
<br />2 X 302 8
<br />MADE
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