• �'1�,' _ l_.6 -cwt, �, .
<br />.GJi' .
<br />0
<br />/ ` V.
<br />,t.6 v L
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<br />TRIGONOMETRIC FORMULtE
<br />B B
<br />c a c a ° a
<br />C A b G�yA: b C
<br />`Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin = ,cos= b ,tan= a , cot = a , see = "Cosec cose= o
<br />c c b a b a
<br />Given Required i z
<br />tan -4=1= b cotB, c= 1 a2 -I z a 1 T
<br />z
<br />a, c A, B, b1 sin A = = cos B, b = %) (c+a (c—a) = c 1— az
<br />p c o
<br />A, a B, A c B=90°—A, b= a cotA, c=
<br />sin A.
<br />A, b B, a, c B=90°—A, a= Ytan A, c= b
<br />f cos A.
<br />A, c B, a, b 'B=90°—A, a -.csinA,:b= ccos A,_
<br />h Solution of Oblique Triangles
<br />Given Required _ a sin B a sin C'
<br />CA, B, a b, c, G' b sin A ' C = 180°—(A -)- B), c = sin _4
<br />. b sin A it sin C
<br />A, a, b. B, c, C sin B= a ,C = 1fi0°—(A -r B), c = sin A
<br />a, b, C A, B, c -4+B=180°-G', tan- (A—B)=(°'—b) ¢n r �A+73),
<br />l _ a sin C +
<br />c sin A
<br />a, b, c A, B, C- s= 2 ,sinzA=ll. b c
<br />(,' a)(s—e)
<br />sin.8= ac C=180* -4.+B)
<br />a, b, c Area s=a f 2+c, area = 1%8(8-2) (s—b) (s—c
<br />A, b;_ e. Area b c sin A.
<br />area = •- 2
<br />A, B, C, a. Area area = a sin B sir. 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 />'Ne Vert. angle=5° 10'. From Table, Page 1%. cos 5- iiy=
<br />c a� n 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />510 P�g1e Horizontal distance also=Slope distance minus slope
<br />distance times (1—cosine of vertical angle). With the
<br />ve same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 50 lo'=.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 It. Horizontal distance=309.6— 14 X 14 =302.6-0.32=302.28 fL
<br />2 X 302.6
<br />WADE IN U.S.A.
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