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5
<br />2—C
<br />r
<br />2,
<br />A"
<br />TRIGONOMETRIC FORMULtE
<br />1Ti B B
<br />G a" c a c a
<br />Ab c A�b C A C
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A, sin = ¢ b a , cot=b —,see= = , cosec =
<br />c c b a b a
<br />Given Required
<br />a,b A,B,c tan A=b= cot B,c= az-} z=a 1+ a2
<br />a, G- A, B, b sin A = =cos B, b = �/ (G (a) (G— a) = G 1 0 l
<br />A, a B, b, c B=90°—A, b= a cotA, c= a
<br />sin A.
<br />A, b B, a, G B = 90°—A, a = b tan A, c = b
<br />cos A.
<br />A, c B, a, b B =900—A, a = c sin A, b = c cos A,
<br />Solution of Oblique Triangles
<br />Given Required a sin B
<br />A, B, a b, c, C b = ' C = 1800—(A + B),,c =asin C
<br />sin A sin A
<br />A, a, b . B c, C sin B= b sin A = 180°—(A B), c =asin C
<br />a C -{- sin A
<br />ao b, C A, B, G A+B=1800— C, tan 1.(A—B)= (a—b) tan z (A+B)
<br />a sin C a +
<br />G =
<br />sin A
<br />a, b, c A, B, C s= 2 ,,A— b e '
<br />sinjB=v I(s—aa(e ° ,C=180°-(A+B)
<br />f' a, b, c Area s = a c , area = s (s—a) (s—b) (s—c
<br />A, b, c Area area = b c 62 A
<br />A, B, C, a Area area - a2 sin B sin C
<br />2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal the
<br />Slope distance multiplied by the
<br />e cosine of the vertical angle. Thus: slope distanceVert. an
<br />=319.4ft.
<br />5 e "sta�c N 9959. Horizontal distance 1fl 4X 9959 318.09 ft,b� 10
<br />lop Angle a Horizontal distance also =Slope distance minus slope
<br />�e�. 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 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=l4 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 IN V. S. A.
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