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<br />TRIGONOMETRIC FORMULAE
<br />B B
<br />a
<br />a a
<br />0 A
<br />C
<br />I Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />-a
<br />For Angle A. �innc 0 a=�,cos=—b ,tan= T b cot = — ,see= cosec
<br />a
<br />Given Requited a � F_ 2
<br />-A B,o V"a2 + I= a IN
<br />a, b tan A = C&B, a +
<br />a2
<br />A, B, b sin = = coo A b =V
<br />T+ aT(c
<br />a,
<br />A, a Al,o B=90°—A, b = a cotA, c=
<br />sin A.
<br />A, b Ae;, a B =90*—A, a = b tan A, e = cos A.
<br />A,, B,' a, b B = 90'—A, a a sin A, b = c dos A,
<br />Solution of Oblique Triangles
<br />Given Req -red asin B asin C
<br />A, B,'d b; 'i C b = sin A , C = 180--(A + B), c = sin A
<br />b sin A a,in C
<br />A, 4, b B, c,'C sinB= a , Of = 180°—(A + B), c = sin A
<br />a, b, 0. A, B, c A+B=180*— 0, tan J(A—B)= (a—b) tan I (A+B)
<br />+ b
<br />a sin C
<br />a sin A
<br />j a, b, a A, B,,C 8=a+b+ JA_—
<br />2,sinbe
<br />a+b+
<br />aI, b, e Area S= 2+ area =N/8(8 a 8 (s —c
<br />basinA
<br />A, b,.c Area. area =
<br />2
<br />a' sin B sin
<br />Area area, = 2 sin A
<br />I REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />cosine of the vertical angle. Thus: slope distance —319-41t
<br />tiQoee Vert. angle= 5Q 101. From Table, Page DL cos 51 IW=
<br />.9959. Horizontal distance=319AX.9959=318.09 ft.
<br />-love 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 50 101=.9959.1—.0959=.0041.
<br />319AX.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 />%ince 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 />2X3n6
<br />+14 MADE IN U. 8. A.
<br />I
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