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<br /> TRIGONOMETRIC FORMULA t
<br /> B B B
<br /> •4
<br /> a aIn
<br /> a
<br /> A
<br /> C b
<br /> Oblique Triangles- C
<br /> Solution of Right Triangles
<br /> a b a b o 0
<br /> ' dor sin = a ,..cos a a ,tan= b,cot=a,see=b, COaec =
<br /> a
<br /> dives lauired
<br /> a,b 'AB,a tanA=b® cotB,c = = a 1T a==
<br /> A, B, b sin A=Q=coe B,b=�/ a+a a—a =0 1—o
<br /> 2 S
<br /> Z, S,. ,Aa B, b, a B=90°—A,b—a cotA,o= sin A.
<br /> G, S A, b B,a, o B=90°—A,a= b tan A,a= cos A.
<br /> G A,c B,a, b I B-900—A,a=a sin A,b=o cos A,
<br /> / z Solution of Oblique Triangles
<br /> S�l Given Required _a sin B
<br /> A, B' a b' c, C b sinA ' C= i80°—(A� �.o = Bin
<br /> b sin A
<br /> ,3, j 4' A, a, b B,o, Q sin B= a ,C= 180°—(A {-B),o = sin A
<br /> '� +
<br /> G° F v a b, C A, B, o A { B=180°—C,tan}(A—B)= a—b ttan+ 8 JB
<br /> 3 ` i c =asin C
<br /> sin A
<br /> v" a+b+aa 8—c
<br /> A, B, C 8= 2 ,sin JA= b a '
<br /> sinjB=� ,C=180°—(A+B)
<br /> c a+b a
<br /> E i a, b, o Area 8=. , area — ) 8—E
<br /> ` 1 j A, b, a Area area = b o sin A
<br /> • � r'` 2
<br /> P
<br /> A,B,C,a Area area =as sin B sin C2 sin A
<br /> a REDUCTION TO HORIZONTAL
<br /> Horizontal distance=Slope distance multiplied by the
<br /> cosine of the vertical angle.Thus:slope distance=819.4 ft
<br /> s�a4� Vert angle=b°104 From Table,Page IX.cos 50 101=
<br /> m 9969. Horizontal distance=319.4X.9969=318.09 ft
<br /> �00g . leHorizontal distance also=Slope distance minus slope
<br /> distance times(1--3osine of vertical angle). With the
<br /> Q same figures as in the preceding example,the follow-
<br /> Horizontal distance inti result is obtained.Cosine b°10'=.9959.1—.9M=.0041.
<br /> 319.4X.0041=1.31.319.4-1.31=31&09 ft
<br /> When the rise is known,the horizontal distance is approximatelWthe elope dist-
<br /> anoe less the square of the rise divided by twice the slope distance. Thus:rise=l4 ft,
<br /> sy1l 8 ft Horizontal disttnoe=3028-2X3M8—=6—&33=80298&
<br /> MACS IN U.S.A.
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