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TRIGONOMETRIC FORMIJUR <br />j: <br />R B <br />a <br />A <br />A <br />C <br />Right Triangle Oblique Triangles <br />Solution of Right, Triangles <br />IVA F�ur Angle A., sin cos tan Cot sec cosec, <br />Given- Required <br />b A. B,e Lan.� =cot Ac = v a + <br />- . .. 1 +�s <br />a, a A. B, b Bin A =coaB, b=N/(o+,7v (a=a) o I <br />02 <br />a <br />20YY. <br />0 B 900—A. b = a cotA, a <br />min A. <br />14- p l t <br />o 5 b <br />j-;, T —Aa =.b tan A <br />A, b J�, -a, c B=90' <br />coo A. <br />A, d B, a, b B-90*—A, a c sin A,b c c os A, <br />Solution of Oblique, Triangles <br />Given Required 'a Bin B a Bin U <br />B a _�A + <br />q.;, A�, 0 1801 <br />ry sin A <br />B, C' C (A+B),o= <br />b sin A asin C <br />b <br />to "I a sin A <br />JV, 211. <br />(a—b) tan (A+B) <br />tan (A—R) � <br />Z b, 0 A, B, c A+B=180* b <br />3 a sin C' <br />sin A <br />A, A 0 8 in JA= <br />2 <br />-C) sinjB=0=180' —(A + R) <br />ac <br />area -1/40 a a .1 C) <br />b, Area <br />2 <br />j <br />A b, c Area b c Bin A <br />2 <br />a, sin B Bill 0 <br />• A, B, C,a Area area 2 sin A <br />12- REDUCTION TO HORIZONTAL <br />Horizontal distance - Slope distance multiplied by the <br />it Ile cosine of the vertical angle. Thus -slope distance -310,4 ft. <br />0 Vert. angle = 50 101. From Table, Page JX; cos 60 IW= <br />oc� ' a 9958. Horizontal d1stanoe=319,4X.99w=3l8_09 ft, <br />'Horizontal distance also- Slope distance minus slope <br />distance times (1 -cosine of vertical angle). With the <br />t same figures as in the preceding example, the follow - <br />.7 Ce ing result is obtained. Cosine 51 10'=.5959.2 -.995[J= -004l. <br />Horizontal distan <br />e7fi When the -rise is known, the horizontal distance is apDroximately:-thi slope dist- <br />ance teas the square of the rise divided by twice the slope distance. Thus: rise=141L, <br />It 14 X 14 ft. <br />slope dLataace=W16 Horizontal distanee=S= 2 X SMG <br />WADE IN U. S. <br />