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r Ste~hCG — <br /> } .TRIGOripmaTRIC FORMUL4•. <br /> 7 o, B' B B <br /> a o a <br /> L7 <br /> A A C <br /> ti C �b C <br /> S'%� 3� �► � » '� Right TnsAgle Oblique Triangles <br /> ` �T S Solution of Right Triangles <br /> a b a b a a <br /> L For Angle$.a rin = o ,cos= a ,tan= b ,Cot= -,sec a =b, cosec= <br /> 1 r _ <br /> Given Wired a �, � <br /> ,o tan A=-_ B,o = a 9 =h <br /> V V b �- 1 + a€ <br /> bo,4��e=eodB,b` by <br /> ,� ✓� ' ,',C; - A,4 a B=90°-A,b =a cotA,o= a <br /> Bin A. , <br /> o B=90°-A,a = b tan A,o= b <br /> `{ cos A. <br /> h A,a. B,a, b B=901-A,a o sin A,b=o coo A, <br /> Solution of Oblique Triangles <br /> lnirea b_a eilt B C r 180 as Bin C <br /> `). z L G 3 sin A ( +' );o = <br /> �' Bin A <br /> .b a. b sin A a Bin C <br /> 3b �- � �` �j,. � 2.:' .�"i7!4 <br /> ,,. a, C sin B= a �C= 180°-(d-f-B):a = <br /> silt <br /> * A+B=180°-C,tan J(A-B)= a-b)tank(, B) <br /> a _asin 0 a+ <br /> sin A <br /> 37 a{-b+o <br /> pu 8-a)(8-c <br /> einB=� a a ,C=180°-(A+4) <br /> 13 0 <br /> b, area <br /> s ; s= 2 , <br /> b, <br /> b o sin A <br /> �e area = 2 <br /> V f"��7 ! I 6 �' a$Bin Bin C '�9 <br /> a Area area = . sin A <br /> 4r, REDUCTION TO HORIZONTAL <br /> Horizontal distance—Slope distance multipu:,! <br /> ' ! cosine of the vertical angle.Thus:slope distma g3" <br /> r a yt� Vert. angle-6 1W. From Table,Page IX Qoa,b 0'— <br /> �" 0 a m 9868..,Herizontal distance-319.4X.986q=SI&0�Y1 <br /> 1e a tancjaf diatance'hlso=Slope distance minim"slope <br /> Jj distance times (1-cosine of vertical angls). W01* the <br /> +� ,� same figurea.as into preceding example,the follows <br /> ©;, f f yl�1 s 6 Z� orizontal distance ing result is obtained Cosine b°101=.81b8.1�.890041. <br /> 319.4X.0041=1.31.318.4-1.31=31&09& <br /> •+per When the rise is]mows,the horlsontal distance is approximately <br /> qpe less the square of the rise divided by twice the slope distance. &I <br /> 6 distance=3029 ft. Florizontai distan"_Mle—a 4 8+302.8—a32--XM28 ft.: <br /> t <br /> IN t <br />