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TRIGONOMETRIC FORMULAE' <br /> B B B <br /> o a e a o <br /> a <br /> Right Ara Oblique Triangles <br /> S Solution of Right Triangles <br /> z 5a b a b o <br /> �3 _4 '..For Angl n= o ,cos— a ,tans h ,cot= a,sec=b, cosec= e <br /> Given a <br /> a, b ,o tanA—!— cotB,o = a 2 —a 1 = <br /> __— -- +a <br /> a <br /> A,a B, b, o B=90o—A,b =a cotA,a= <br /> sin A. <br /> ate. B,a, a B=900—A,a =b tan A,a= b <br /> coe A. <br /> " �A,c B, a, b I B=90'—A,a=o sin A,b=o cos A, <br /> Solution of Oblique Triangles <br /> *m a sin B <br /> Ba <br /> ' 1 C b __ sin A ' C= 180o—(A+B),o =asin C <br /> sin A <br /> C sin B= b sa A,C= 180°—(A+B),o = a ein C <br /> sin A <br /> A+B=180'—C,tan j(A-B)— a—b tan (A+B). <br /> �- �'%G' _asin C a+ b <br /> % ._ ° sinA <br /> b, a , C -aa+b+ � =n A— <br /> z. A s 2 b o <br /> a einjB JB—a a—° C_180°—{A+B) <br /> .•, V a o <br /> a+b+a <br /> y k yj a, b, o 8= 2 , area = s+—a a— e—c <br /> basinA <br /> u A, b, o • area = 2 <br /> `ter <br /> A B C s Atie area Q ai sin B sin C <br /> ' 2 sin A <br /> `I REDUCTION TO HORIZONTAL <br /> l _ Horizontal distance—Slope distance multiplied by the <br /> cosineoftheveoicel0*1e,1'6omslope distance-M4ft. <br /> Vert angle=6 1or. From 1'a c <br /> 8868. Horizontal distance—Meb4,X.9e% i1�8.68�6 loam , <br /> 5� Horizontal distance also=may distance Minns slops <br /> distance times 11--ooslna of Irtloai With the <br /> samee res as in the p the.tolkn►- < <br /> l Horizontal dfitance ine , is obtained.Cosine 6°1�Y 1—.8868=.OkL ;. <br /> When the rise is Itaoree,the hor$rantal dis�oe is pa &� slgpe d1H <br /> ance less the square of the ries divided by twiesthe slope distanoe. risi1 14 ft. <br /> slope distance-102.8 ft Horizontal diad no*-M6-- X 14 <br /> 4X Me <br /> AIM <br /> • <br /> r , <br /> Y <br /> • e <br /> .. X <br /> s <br />