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<br />. - �T I�G(�NOMRTR ICF ORMULiE. Sa flip,
<br />j��—= Il e• B
<br />01
<br />c a c c
<br />n. a
<br />'A A '
<br />t b C �b C b ► C
<br />Right Triangle Oblique Triangles
<br />i Solution of Right Triangles
<br />For Angle A.: sin = 2 , cos h ton= a , cot = b , sec = cosec = e
<br />a b' a
<br />Given Required
<br />a, b _9, BM,c tan A= b = cot B, c = a ++ z = a 1
<br />a
<br />ti.
<br />a, c A, B; b sin A'= a =cos B, b = �/-}-
<br />(ca) (c—a) = c .� 1 —a
<br />�
<br />C. . 02 q!"
<br />A,'a B, _b,' cB=90°—A,b=acotA,c= a o 5 IF.
<br />�'
<br />sin A. / a
<br />A, b B, a c B=90°—A,a = btan A,c= b 3
<br />cos A. - 6-c `
<br />A, c B=90'—A, a = c sin A, b= c cosA, _ 7 v
<br />3`
<br />Solution of Oblique Triangles' �'% :t
<br />Given -Requireda sin B
<br />_
<br />A, B, a b, o; C b sin A ' C = 180°—(A + B), c = asin C
<br />sin A
<br />b sin -Ary
<br />i = a ,C ='180°—(A + ), c = a
<br />in
<br />�, a, b L', c,. C sin BB
<br />-
<br />sin A �?
<br />a, b, C A, B; c 4 B- a—b) tan z (A+B '
<br />f + —180 — C, tan � (�—B)— a
<br />+ b �'
<br />a sin C
<br />i7 c sin A. f 32-4
<br />a, b, c A, BG s=a+b+
<br />, c,sin3` —d— 1ls-b)(s—c)
<br />t 66 �� 6� sin iB= _ 3 4-o
<br />fi✓ a cd+B0�2 190--4) ,
<br />�t '
<br />38..0
<br />-.. Area s _ 2 area = s(.r—a (s— ) (s—c
<br />.,
<br />I!. .Qs,•b; c � Area- area = b c sin -4
<br />a'- sin B sin C
<br />A,B,.C,a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL.
<br />Horizontal distance = Slope distance multiplied by the
<br />e cosineofthe verticalangle.Thus: slopedistance=319.4ft.
<br />e als�anc Vert. angle=5° 101. From Table, Page IX. cos 50 101=
<br />9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />r 5�Op �n'Ne xti Horizontal distance also=Slone distance minus slope
<br />Nett. 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 5°101=.9259.1—.9959=.0041.
<br />j319.4X.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.,f
<br />slope distance=302.6 ft. Horizontal distance=3o2.6— 14 X 14 =302.6-0.32=302.23ft.
<br />2 X 302.6 _
<br />�' • MhDE IM U. S.A.
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