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yT <br />est S! J Ja/6 M/ <br />vS i <br />TRIGONOMETRIC. FORMU i <br />o p <br />a <br />I g CA <br />Right Triangle Oblique. Triangles <br />l Solution of Right 'Triangles — — <br />u� <br />For Angie A. sin = , cos c , tan= b , cot a , sec ^ b , cosec o <br />a <br />Given Required a z <br />a, b A, B ,c tan A = b = cot B,:a az = a I'+ <br />a2 <br />77 z <br />C A, B, b a <br />S, v <br />/ ' <br />� A, a B,_ b, c 8=90°—A, b = acotA, c=' a <br />A, b ' B, a, G B = 90°—A, a _ b tan A, c= <br />b <br />i I cos A. <br />° <br />c B, a, b B-90— a = e siu A, b = c cos A., <br />Solution of Oblique Triangles <br />Given Rdiauired a sin B a sin C <br />A B, a b, q C b sin A ; C = 180`--(A j B), G = <br />A <br />stn <br />A, a, b B, c, G' sin B = bsin A a <br />, Q sin C = 180°—(A } B}, c = <br />a sin A <br />C A, B, e A+B=180°— C, tan (A—B)— (a—b) a ' �a +B), <br />( <br />l' 4' a' <br />a sin C <br />• <br />_ <br />" a ein A <br />b, a A, B, C s—a+b+a,sin?A—AI (s— <br />b)(a—c <br />sin JB={s—a4% C=180°—(A-+-B) <br />1 <br />a+b+c <br />te, b, c Area e = 2 ,area <br />A, b, c Areab a sin A <br />Brea <br />2 <br />as sin B sin 4 <br />4, B, C, a Area Brea = <br />2 sin A <br />REDUCTION TO HORIZONTAL <br />Horizontal distance= Slope distance multiplied by the <br />E <br />e cosine oftheverLcalangle.Thus- slope distance =319.4 ft. <br />1 d+s 'e Vert. angle= 511Io'. From Table, Page 1Y_ cos 50 1N= <br />9959. Horizontal distance=319.4X.9959=318.09 ft. <br />aQ e w Horizontal distance also=Slope distance minus slope <br />atie <br />! distance <br />ve times (1—cosine of vertical angle). With the <br />r <br />same fiP ores as in the preceding example, the follow - <br />Horizontal distance Ing result is obtained. Cosine 5°t0A=.9959.1—.9959=.0041. <br />319.4X.0041=1.31. 319.4-1.31=313.09 ft. <br />When the -rise is known, the horizontal distance is approximately:—the slope dist- <br />once less the square of the rise divided by twice the slope distance. Thus: rise=l4 ft, <br />t <br />slope distanee=302B ft Harizontal distance=3028— 14 X 14 =3028-0.32=30223 fL <br />! <br />2X3028 <br />MADE IN u. 6-A. <br />- - <br />