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<br />TRIGONOMETRIC FORMULeE
<br />!c a c a c a
<br />�.
<br />i
<br />A�b
<br />i'Right Triangle. Oblique Triangles
<br />Solution of Right Triangles
<br />a b a b a c
<br />For Angle A. sin = e , cos = c , tan = , cbt = a , see = b , cosec = a
<br />Given - Required z
<br />a,b A,B,c tan A=�--cotB,a= az+bz=a .1+'(Z
<br />R
<br />{ a,0 A,B,b sin A=C=cos B,b=�6(a o—a)=c1—`per
<br />a
<br />A, a.
<br />A, b
<br />A, c
<br />Given
<br />A, B, a
<br />A,a,b-
<br />B, b, c
<br />B=90°—A, b = a cotA, c= sin A.
<br />l%
<br />b
<br />B, a, c
<br />B = 90°—A, a = b tan A, c =
<br />cos A.
<br />B, a, b
<br />B=90°—A,a=cSin A,b=eco6A,
<br />�a
<br />gara:9�sat�•.R
<br />r
<br />�aaSGn,
<br />~ }'•G
<br />°�
<br />��iQ 60 -5-
<br />P1 i�gg+t%'4!
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<br />B, c, C'
<br />. b sinA asin C
<br />sin B = , C = 180°—(A -i- B), o =
<br />p^ ,,•
<br />a sin A
<br />d
<br />TRIGONOMETRIC FORMULeE
<br />!c a c a c a
<br />�.
<br />i
<br />A�b
<br />i'Right Triangle. Oblique Triangles
<br />Solution of Right Triangles
<br />a b a b a c
<br />For Angle A. sin = e , cos = c , tan = , cbt = a , see = b , cosec = a
<br />Given - Required z
<br />a,b A,B,c tan A=�--cotB,a= az+bz=a .1+'(Z
<br />R
<br />{ a,0 A,B,b sin A=C=cos B,b=�6(a o—a)=c1—`per
<br />a
<br />A, a.
<br />A, b
<br />A, c
<br />Given
<br />A, B, a
<br />A,a,b-
<br />B, b, c
<br />B=90°—A, b = a cotA, c= sin A.
<br />l%
<br />b
<br />B, a, c
<br />B = 90°—A, a = b tan A, c =
<br />cos A.
<br />B, a, b
<br />B=90°—A,a=cSin A,b=eco6A,
<br />�a
<br />Solution of Oblique Triangles '-
<br />r
<br />Requiredj'
<br />' C = 180°—(A -1- B),'c =
<br />sin A aiii r1-
<br />��-
<br />B, c, C'
<br />. b sinA asin C
<br />sin B = , C = 180°—(A -i- B), o =
<br />p^ ,,•
<br />a sin A
<br />d
<br />(a—b) tan '
<br />b, a A, B, c A;, B -180°—C, tan J(A—B)=
<br />a sin C +
<br />c= sinA
<br />ds, b, c A, B, C s—a++c,sinA—�{)
<br />i
<br />be
<br />sin JB=Aa(`c ) , C=180°—(A+B)
<br />ia, b, c Area s= 2 , area
<br />( A, b, c Area area = b c aim A
<br />t 2
<br />a4 sin B sin a
<br />A, B, C,a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied bs, the
<br />cosine of the vertical angle. Thus: slope distance r310.4 ft.
<br />tQfice Vert. ariale=5110'. From Table, Page IX. cos 5° 10'=
<br />I: 3�S m 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />!P P 5,ope Age distan ettimestU—cosine ofIsG= Slope verticalangle).distance 1slope
<br />e same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 50 10'=.9959. 1—.9959=.0041.
<br />319.4><.0041=1.31.319.4-1.81=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. HorizontaI distance=3026— 14 X 14 =502,9-0.32=902.28ft.
<br />2X=6
<br />MADE IN U.S.A.
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