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<br />62
<br />TRIGONOMETRIC FORMULiE
<br />v>
<br />2- 6 n n 3 0 B B
<br />&0 3vGG 3
<br />381 .4 G `__— /"� c a ° a c a
<br />✓ O
<br />C A b C A b C
<br />t
<br />Right Triangle Oblique Triangles-
<br />' Solution of Right Triangles
<br />S
<br />'For Angle A. sin = a , cos = a , tan= b , cot = a , see = b , cosec = a'
<br />7 g Given Required
<br />a,b A,B,c tanA=b=cotB,c= a2+-{ az
<br />0
<br />% �� a, 0 A, B, b sin A = — =cos B, b = c a c—a
<br />¢ 0 3 1 \/(—+� )=o�1—;—
<br />J
<br />Z.S 8 7 I6 A, a B, b,. c B=90°—A, b = a cotA, c= sin A.
<br />)�,`� Z 0/5(�
<br />46 •3 7 2 Ff`� b ✓�
<br />` ? A, b B, a, c B = 90°—A, a = b tan'A, c = cos A.
<br />1 3 (0 5 O S f A, r, 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 />l.( �•��� j - 51' d, B,a b, c„C b— sin A'C=180°—(A+B),c= sin
<br />Dl ;
<br />10\1° A,a,b B,o,C sin B=bs2A,C=180°—(A+B),c=asinC
<br />dJ i sin A
<br />b, C A, B, c A+B=180°— C, tan 1.(A—B)= (a—b) tan z (A+B) .
<br />a + b
<br />a sin C
<br />c sin A
<br />4 4 (09 a+b i
<br />Z Cal (p. 4— S a, b, o A, B, C s = 2 "in 2A=� be
<br />,
<br />it .,
<br />sin21B=),C=180°—(A+B)
<br />�ZI a+b+c
<br />a b, c Area s = 2 ,area = s(s—a s— ) (.1 1)
<br />3 p g Z C7 _! 6 ¢� 9 l � A, b, c Area area = b c sin A
<br />0.% r Oji 2 a2 sin B sin C
<br />A, B, Q a Area area = 2 sin A
<br />$ 5.8 4— s y REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied by the
<br />0 S iJ Z- cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />'T l �, Q tsnce Vert. angle =5° 101. From Table, Page IX. cos 5° 10l=
<br />b 61 d 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />610 c, ArSW Horizontal distance also—Slope distance minus slope
<br />_ e a distance times (1—cosine of vertical angle). With the
<br />A Z �, 1 �P C / O D a �l �. �i same figures as in the preceding example, the follow -
<br />`f” ( 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 />�, 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 />4 0 g 7 7 L slope distance=302.6 ft. Horizontal distance=302.6— 14 X 14 =3026-0.32=302.28 ft.
<br />3� 1 2 X 302.6
<br />.- MADE W U. & Ai
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