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TRIGONOMETRIC FORMULfE
<br />B I3 B
<br />a a a
<br />A' -
<br />b 'C A.� b• C A b C
<br />Right Triangle I Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. • sin = ,cos = b ,tan = a ,cot = b ,sec = a , cosec =
<br />c c b a b a
<br />Given Required
<br />a, IV A,-B;c tan A=b=cot Ae N/—al-772 = a 1 f az
<br />a, c A, B, b sin A = c = cos B, b = �/ o -{-a (c—a) = c J 1— a'
<br />i
<br />A,,a • .B,'b; c' ' B=90°—A, b = a cotA, o= a
<br />sin A.
<br />A, b. B, a, c B=90 *—A;a = b tan A, c = b
<br />cos A.
<br />A, c B, a, b B = 90°=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, G' b = sin A , C = 180°—(A + B), o =asin O
<br />sin A
<br />' b sin A
<br />A, a;.b B, c, C sin B= a ,C=180°—(A+B),c=asinC
<br />sin A
<br />a, b, C A, B, e, A+B=1800— C, tan ; (A—B)= (a—b) tan z (A+ B)
<br />ab
<br />asin C +
<br />sin A
<br />•a, b, c A, B, C s=a+b+0,sin` — A=
<br />2 N b c
<br />sin JB= (s a o ,C=180°—(A+B)
<br />a, b, c Area 4=a+2+c, area
<br />A, b, c Areab e sin A
<br />area = 2
<br />a? sin B sin C
<br />A; Bi Q, a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />4J Horizontal distance= Slope distance multiplied by the
<br />cosinaree
<br />Vert. angle ng ee vertical5° 101. From able slope
<br />P gelIXncos 5° 10Yce =319.4 =�
<br />#' oQe 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />S� Apg�e a Horizontal distance also=Slope distance minus slope
<br />4e 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—.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 302.8
<br />- MADE Ia Y. &A.
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