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�rt1r- °—, Baa=__L..61�11_:''
<br />TRIGONOMETRIC FORMULAE
<br />B B
<br />�3 3q `' A A A C
<br />�bC b
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />�y, _ cc b a b c o
<br />a ( For Angle A; ein = .c ,cos,= c ;tan = b ,cot = a ,sec = b , cosec = a
<br />.� Given Required a a
<br />(� p•� ` a,b• A;'B,c tanA=b=cotB,c= a2-{- 2=a 1 } 2a
<br />a, e. A; B, b sin A = : = cos B, b = N/ (c+a) (c—a = c 1—Q e
<br />A,'a• B, b. c- B=90°-�, b = a cotA, o= a
<br />,
<br />sin A,
<br />B, a, c B=9Q—A,a = btan A,c=
<br />cos A.
<br />e B, a, b B=90* A,a = c sin A, b = c cos A,
<br />Solution of. Oblique Triangles
<br />Given' Required
<br />B a, b c C b` a sin B C= 180°—QA 4 B) a= a sin C
<br />B, C '
<br />sin A ' sin A
<br />b sin A a sin C
<br />B C 180°—(A B),
<br />Z
<br />A, a,, b 0,
<br />,�
<br />sin = , = + c =
<br />a sinA
<br />c(
<br />p J _
<br />tea, Z (
<br />a, b, C.; A, B, c
<br />( ) 1(+ )
<br />A+B=180°— C, tan I (A—B)= a -b tan i ( B
<br />k
<br />�4
<br />a sin V
<br />sin A
<br />r�
<br />c
<br />-©�%
<br />-
<br />a;'b, a A, B, C
<br />a b
<br />+ + 1 c—)�
<br />s= ,sinzA=� ,
<br />S'c'
<br />�
<br />�.t
<br />sin �B= I9 Q�� ),C=180°—(A+B)
<br />s1,�
<br />J
<br />�
<br />1 21 �l ` '
<br />(---
<br />_
<br />¢ 1t :) j
<br />.A4' a, b, cac
<br />Area
<br />8= + + ,area
<br />o a n- . a
<br />—^
<br />__— _�b
<br />'Area
<br />a sin A
<br />area =
<br />2
<br />J�
<br />Area
<br />2 in B sin
<br />area = a s
<br />2 sin A C -
<br />C7
<br />REDUCTION TO HORIZONTAL
<br />j �_ ,�
<br />3 '''
<br />Horizontal distance—Slope distance multiplied by the
<br />cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />Vert. From Table,
<br />1
<br />-� �'�
<br />/
<br />z Star°e
<br />a
<br />angle =5° IV. Page IX. cos 51 101=
<br />9959. Horizontal distance= X ft
<br />,ove ,e
<br />p¢
<br />lop
<br />Horizontal distance also= Slope distance minus slope
<br />distance mi
<br />distance times (1—cosine of vertical angle). With the
<br />}
<br />eS�ing
<br />same figures as in the preceding example; the follow-
<br />Horizontal distance result is obtained. Cosine 50 101=.9959.1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=318.09 ft.
<br />When the rise is known,
<br />the horizontal distance is approximately:—tbe slope dist-
<br />]
<br />ante less the square of the
<br />rise divided by twice the slope distance. Thus: rise=14 ft.,
<br />/ \1'4X14—
<br />X
<br />slope distance=3026 ft.
<br />Horizontal distance=3026— , —3026-0.32=302.28 ft
<br />/ 2X302.6
<br />i
<br />' NADE IaY. 6. Al
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