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ria __ � j ,� . .. • <br /> r .,� lb•71 - . <br /> r df d <br /> 1"if t T a Oblique Trlan6es 1 <br /> _ J Satiort liisl:t rlunles Y <br /> a a <br /> 3 y Z For Angle kiln = ,'cos *o 'ten b'co►— a' "3► y1*s� <br /> Give <br /> Z�. <br /> -Z , 4 <br /> tan.e).b=cot B,o <br /> AnAmQ cos B,b 0-}•4 0--a <br /> 3 7 ,a a B•80°-d,b m,a cotd,o g 0sia A. <br /> A — _ <br /> b ,��ty a 8+.90° A,a b tan A,e b <br /> -- — cos A' 3 <br /> A,c b B=W,—A,a =csinA,b= DcosA, <br /> t = 7 Solution of Oblique Triangles <br /> Givm �dasinB <br /> ._ A, A.a b, dI,'.0 b = cin A ' C= 180°—(A+B),o= cin d <br /> A �4 C sin B= b sin A C= 1800—(A+B),o a eia C <br /> a sin A + <br /> (a—b)tan,}(A+B) <br /> S_ j ,� asb, C ; B,c A+B=180°—C,tan a(A—B) a+ b �. <br /> a sin C <br /> _ � . F —� •. sin <br /> (,' a=a+Z+OSin'A=17 la be—c <br /> sinB= I(e—axa—c ,C=1806—(A+B) <br /> a c <br /> > 6 .� <br /> a=a+b+o, rea a—c <br /> r <br /> 3 <br /> b e s2 A <br /> area = <br /> a'sin B sin C <br /> q / area = 2 sin A <br /> REDUCTION TO HORIZONTAL <br /> ' - Horizontal distance=Slope cemultiplledbythe <br /> 1 S ? > , '.` cosine of thevertioalangl. h, dopedistaace=81&4fL <br /> :� " f— e Vertangle=6°ld. From Ta Pa`e-I3�ccs 60101= ` <br /> s — = <br /> 0,.%69. Horizontal distan((, 81 9969 1&99 tt d <br /> Horizontal distance also=,Slo ,distance uiipns al� ± <br /> 3 distance times U--cosine Of vetti�i ). With <br /> same figures as in a pre0edina K the tgnow- <br /> -- ing result is obtained.Cosine 6°.N9.1�—.9Y66=441E <br /> Horizon ce 8194X.0041=1.81.819.4-1.81181& ft <br /> > s Wbe is known,the horizontal distance is app el -4he slop , <br /> of the rise divided by twice the slope distance..Thua e3te-14 , <br /> i s Ope s d—lJX JA <br /> l ... y. r w•w �. .,.•.,w� NAD!IMFC16 <br /> • t, <br />