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TRIGONOMETRIC FORMULAE
<br /> _ B B B
<br /> ,o c
<br /> 3ik-r g 1 a c a ° a
<br /> qx� �
<br /> b o A b C
<br /> R A b C
<br /> ight Triangle �--Oblique Triangles-
<br /> Solution of Right Triangles
<br /> For Angle A. sin = a ,cos= b,tan= a ,cot = b ,sec=�, cosec= a
<br /> i Given Required I C c b a b �y
<br /> A, B,c tan A=a= cot c =B, a z =
<br /> b 2+ a I T a
<br /> a
<br /> A,B, b
<br /> A,a B, b,'c B=90°—A, b =acotA,c= a
<br /> sin A.
<br /> A, b B, a, c B=90'—A,a = b tan A,c= b
<br /> cos A.
<br /> f A,c B, a, b I B=90°—A,a =cSin A,b= ecos A,
<br /> Solution of Oblique Triangles
<br /> <- f Given Required a sin B
<br /> ' A' B,a b' e, C b = sin A ' C= 180`—(A+B), c — a,in C
<br /> sin A
<br /> b sin A
<br /> a, b B,c, C sin B= ,C= 180°—(A-t-B),c = a sinC
<br /> a sin A
<br /> C A, B,c A+B=180°—C,tan (A—B)— a—b)tan- (A+B)
<br /> 7j 7b` D S c = inC a+ b
<br /> as
<br /> q i J sin A
<br /> ay b, c A, B, C s=a+b+c ��"�)
<br /> 2 ,six aA=
<br /> sinlB=�(y ac ,C=180--(A+B)
<br /> 9 a+b+c
<br /> t7 b, c Area 8= 2 , area =V/a(s—a s— (s—o
<br /> a Areab e sin A
<br /> area =
<br /> 2
<br /> A,B,C,a Area area =az sin B sin C 2 sin A
<br /> \ REDUCTION TO HORIZONTAL
<br /> Horizontal distance—Slope distance multiplied by the
<br /> cosine of the vertical angle.Thus:slope distance=s19.4 rL
<br /> pee Vert. angle=60 1W. From Table,Page IX.cos 6o 1W=
<br /> a, 996& Horizontal distahce=519.4X.9M=sl&09 r1.
<br /> 6�uoe brgle ex Horizontal distance also=Slope distance minus alo
<br /> Qe 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 501N=.9968,i—AW=.0041.
<br /> 319.4X.0041=1.31.919.4-1.31=31&O9 ft.
<br /> dist-
<br /> ance
<br /> When
<br /> squarrise e known,
<br /> f a rise divided by twiceal the slop approximately:—the
<br /> Th neislope
<br /> oe 14 ft,
<br /> slope distance--=@ f1. Ei4ritontal distar�pa=bps s-lf.2s.l�.�pyg_0 3��48�
<br /> _ 2 X SMS
<br /> MADS M U.s,A.
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