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VII&I
<br />.J 6; t1 6, cf
<br />13 U-
<br />iG t4 ;.
<br />7 yam.- h-
<br />t 'ate •.
<br />Vin, 7 131A ?.`' �` ✓ r?,
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<br />n � e
<br />TRIGONOMETRIC FORMULtE
<br />A z•'° 4 35 �i . A
<br />b C b G' b C
<br />Right,.T_ rianglc Oblique Triangles .
<br />Solution of Right 'Triangles
<br />a b a b c
<br />For Angle A. sin = c ,cos= c , Laa= b ,cot = �., scc = b, cosec =
<br />a
<br />Given Required
<br />a, b _4, B ,c tan A = b =- cot B, c=.1/ct= +2�= a 1 + 4
<br />_ a
<br />a, c� A., B, b- sin A = c = cos B, b = \1(c=, a) (c—a) =CAI I-
<br />0 9
<br />.b
<br />A, a B; a b, c J3 90°-A, b = a cotA, c = sin A.
<br />A,b B, a, e B=90°—A,a=ab tan A, c= b
<br />cos A.
<br />A, c B, a,, b 13 = 90°_=A, a = o sin A, b = c cos A, y"
<br />`:Solution of Oblique Triangles �'• .
<br />Given Required •I
<br />A, B, a L e,, C b ' sinn ' C = 180°—(A + B), c.= sin A
<br />b'sin A a sin C
<br />A, a, b B, r., C sin B.= a , C = 180°—(A t.B), e = sin A
<br />a, b, C A, B, c A.+B=180°--C, tan ' (A—B)= (a—b) ten (A+B)
<br />a + b
<br />a sin C.
<br />e — sinA 11-4o
<br />a, b,c. A, B, G s= 2 ,sin 3 l= J b a �u f
<br />sin,'B=A B—`1('—all'—c) , C=180°—(A+B)
<br />ac'
<br />a+b+c
<br />a, .b, a Area s= 2 ,area (s --c)
<br />A, b, c Area. aiea = b c sin' A
<br />2
<br />A B C, a 'Area
<br />a2 sin B sin C
<br />' 2 sin A
<br />REDUCTION TO HORIZO
<br />Horizontal distance =Slope ermultipiied by the
<br />cosineofthecerticalangle. o"pe distance=37,9.4 ft f
<br />farce Vert. angle=5° 10'." Frb' !Page IX. eos 6� 20�=;�.
<br />e ass` ! .9959. Horizontal distant y .9959=318.09 ft.
<br />Horizontal distance in11
<br />194X.001
<br />When the rise is known, the horizo
<br />ante less the square of the rise divided
<br />slope distance=3026 ft Horizontal dis
<br />
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