Laserfiche WebLink
TRIGONOMETRIC FORMULj9 <br />rB It <br />B <br />a <br />a <br />T4 —40. A A. <br />413 - b <br />Right Triangle Oblique Triangles <br />Solution of Right Triangles <br />For Angle A. sin = a cos = b , tan= a I cot b sec cosec <br />a <br />Given" Required _F2 <br />a� A, B,o tan. <br />A=!= cotB o a <br />4-' b <br />+77 <br />2 <br />a; <br />0. <br />A, B, b Bin A cos B, b (e+a7(c—a) o — <br />0. a <br />b, c B=90°—A, b a cot <br />v7 i'A, B, <br />sin A. <br />.1A, b B, a, e B =90*—A. a b tan A, c <br />a <br />c.. A. C) <br />as A, c A a, b I B 90'—A, a a sin A, b c cos A, <br />Solution of Oblique Triangles <br />Given Required asinB in C <br />A, B, a b, e, C C = 1:80'—(A + B), <br />Bin A sin A <br />b sin Aa, C Aa sin C <br />--(A + B) , c = <br />sin a , C = 180 sin A <br />b, C A, B, a A+B=180*— 0, tan j (A—B)_ (a—b) ta '(A+B) <br />a + <br />71, a sin C <br />sin A <br />b + <br />b, A, B, C 8=a+Bin JA= <br />2 N s be <br />sin jB= C= 180._ <br />N ae (A+B) <br />a+b+e. <br />—0-7 a, b, c Area 8 = 2 area = s (8__7T8 <br />A, b, c Areab.ciin A <br />2 <br />a' sin B sin C <br />a Area area = <br />2 sin A <br />REDUCTION TO HORIZONTAL <br />Horizontal distance=Slope distance multiplied by the <br />ce cosine of the vertical angle. Thus: slope distance =319.4 ft. angle — 50 101. From Table, Page IX cos 50 101�' <br />e.Horizontal distance 9m. Horizontal distance=319AX.9959=318.09 ft. <br />li <br />le .ri. di'-Iso=Slove distance minus slope <br />distance times (1—cosine of vertical angle). With the <br />same figures as in the preceding example, the follow- <br />Horiiontal distance ing result is obtained. Cosine 51101=.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 />pace less less the square of the rise divided by twice the slope distance. Thus: rise=14M. <br />-h —302 6 ft. Horizontal distance==6— 14 X 14 =3046--0A2=302.28 ft. <br />2 X 302.6 <br />MADE IN U. S. A. <br />