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�� <br />��� <br />,�•�'<�/L � <br />, , ,`• - '. TRIGONO.METRIC FORMULfE <br />II i, <br />sl�Q�S <br />B 11 B <br />c' <br />/ 4-7 <br />OJ, / <br />a'A <br />a c a a <br />AA b CA b C <br />b C <br />1 <br />- <br />Right Oblique Triangles <br />_Triangle <br />Solutiori of Right Triangles <br />,f <br />For Angle A. sin = a ,cos= b ,tan= a ,cot = b ,sec = �, cosec = c <br />b b a <br />I <br />c c a <br />o <br />.. <br />/ <br />Given Required a <br />a,b A,B,c tanA=b=cot.R,c= al } ==a 1+ <br />.. - — „ <br />z <br />a, c 'A, B, b sin A = = cos-B, b =,V(c+a) (c—a) = c 1 — o <br />'fly" <br />1 - �". <br />,�r✓��'L.. <br />(� <br />a <br />A; a B, b, c B=90°—A, b = a cotd, c= <br />sin A. <br />c <br />b <br />d, b B, a, c B = 90°—A, a = b tan A, c = <br />cos d. <br />3 r t <br />A, c B, a, b B = 90°—A, a = c sin A, &= c cos-4, ^ <br />l <br />Solution of Oblique Triangles <br />— <br />! Given Required a sin B ' a sin C <br />7 A, B, a b, c, C b = <br />A ' C = 1801—(A + B), c = sin A <br />sin <br />utb <br />sin A G a sill C <br />A, a, b B, 0,C sin B= C = 180°—(A + B), c = <br />d <br />a, sin <br />1 <br />a, b, C A, B, e, A+B=180°—C, tan :'(A—B)=(a—b) tan 1(d+B) <br />t <br />a + b <br />�,'-. <br />a sin C <br />J <br />I <br />/ p <br /><O(DI�� <br />I _ - <br />c s in _4 <br />2, 2i <br />1 <br />SDI, <br />- <br />a+b+c ls- b)(—c) <br />b,. c A; B, C S= 2 ,Sin-2A= 1 b c <br />; <br />g <br />sin' B= Il' , C=180°—(A +B) <br />1 (L C <br />A <br />/ <br />('t.�S ' <br />�' 63--- <br />a, b, c Area s=a+2 ! c, area <br />li <br />"� <br />bosinA <br />A, b, c . .Area area <br />,. <br />�+w�_~.� <br />2 <br />LrD <br />a= sin B sin C <br />499 <br />%, �� .. <br />A, B, C, a Area area = <br />2 sin A <br />REDUCTION TO HORIZONTAL <br />Horizontal distance=Slope distance multiplied by the <br />cosine oft ie vertical angle. Thus: slope distance =319.4 ft. <br />Vert. angle=51 101. From Table, Page IX. cos 5° 11Y= <br />' <br />''<, <br />9959. Horizontal distance=319.4X.9959=318.09 ft. <br />s�opc P��le Horizontal distance also=Slope distance minus slope <br />ye distance times (1—cosine of vertical angle). With the <br />/ <br />��t 06 <br />-7q,06 <br />'f' �? <br />ff 1 <br />ollow- <br />same figures as in the preceding example, the follow- <br />Horizontal distance ing result is obtained- Cosine 50 101=•9959.1—.9959=.0041. <br />319.4X.0041=1.31. 319.4-1.31=318.09 ft. <br />When the rise is known, the horizontal distance is approximately:—the slope dist- <br />67.05 <br />° <br />ante less the square of the rise divided by twice the slope distance. Thus: rise=141t., <br />14 X 14 <br />rf � <br />S � � <br />slope distance=302.6 ft. Horizontal distance=302.6— =302 6-0.32=302.28 ft. <br />2 X 302.6 • <br />16 <br />MADE IN U. &A. <br />L <br />,moi <br />