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TRIGONOMETRIC F.ORMUL& l
<br />B B B
<br />a
<br />r
<br />Right Triangle Oblique .Triangles.
<br />j Solution of Right Triangles'
<br />), For Angle A. sin= a cos =- b ,. tan = a , cot = b , sec = cosec = a 5
<br />a b., a
<br />! Given Required
<br />a2
<br />a,b A,B,c tanA=b=cotB,c= a2 a=a '1-f
<br />2
<br />o A, B, b sin A = a = cos B, b = as
<br />A, - a B, b, c B=90°—A, b= a cotA, o= a _
<br />sin A.
<br />A, b-- B, a, a 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
<br />A, B,a b, c, C b= snA'C=180°—(A+B),c= sin
<br />b sin A a sin C
<br />A, a, b- B, e, C sin B=, a ,C = 1800—(A (B), e = sin A
<br />a, b, C A, B, c A+B--180°— C, tan 2' (A—B)- (a—b) tan z (A+B)�
<br />_asin C a+b
<br />c
<br />sin A '
<br />a -b f o I(s— b)(s—c)
<br />a, b, c A, B, C s=2 ,sin' -,A=., b c '
<br />sin %B=l(s—a)(c ) , C= 180°—(� +B)
<br />a, b, c. Area 8=a+2+e,.area
<br />• A, b, c Area area = besin A 2
<br />aE sin B sin C
<br />A, B, C, a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Y Horizontal distance=Slope distance multiplied by the
<br />cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />�aoce Vert. angle=51101. From Table, Page IX. cos 50101=
<br />e als y 9959. Horizontal distance -319.4X.9959=318.09 ft.
<br />sop 11191 distancHorizoe timest(lcco (1—cosine oflvertcal angle).minus
<br />Withlthe
<br />ve same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 50 10f=.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 302.6
<br />MADE IN U. B. A.
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