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TRIGONOMETRIC FORMUL,1E
<br />/'til •�✓ sa g''- B B
<br />�t u a ° a c a
<br />`. b C f C
<br />1Rigbt Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />i a b a b c
<br />For Angle A. sin = c , cos = c , tan= b , cot = a , sec = b cosec = u
<br />Given Required =
<br />a, b: A, B ,c tan A = b = cot B, e = �y2 + 2 = a 1 + as
<br />e
<br />A, B, b
<br />A, a B, b, c B=90°—A, b= a cotA, c= a .
<br />sin A.
<br />B, a, c B = 90°-4, a = b tan A, c =. cos A.
<br />c_ B, a, b B = 90°—A, a = c sin A, b = c cos A,�`yi
<br />Solution of Oblique Triangles
<br />Given Required
<br />A,'B,a b, c; C b- snA'C=180°—(A+•B),c= sin
<br />b sin A a sin C
<br />a, b B, c, C sing= a ,C = 180°—(A } B), c = sin A
<br />r
<br />(a—b) tan !,.(A+B)
<br />b, .0 A, B, o A+B=180°—.0, tan ; (A—B)= a ,
<br />_ a sin C
<br />1 r 6 sin A'..
<br />' a+b+c , .I s --b) s—o
<br />b, a A, B, C 8= 2 ,sin 7A =-Y b c ,
<br />sin JB=,\17 a(c C=180°=(d.+B)
<br />a+b+c
<br />6s, b,• c • Area s= 2 , area = s(s—a) s— (s ---c
<br />b. c sin A
<br />t, A; k c" Area area =
<br />as sin B sin C
<br />A, B, C, a 'Area ares = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance =Slope distance multiplied by the
<br />cosine of the vertical angle. Thus: slope distance =319.4tt.
<br />dist�4ce. Vert. angle=5° 101. From Table, Page IX. cos 5°:10'•-
<br />9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />r�104e Ne
<br />4 Horizontal distance also= Slope distance minus slope
<br />distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 50 101=.9959.1—.9959=.0011.
<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 />ante less the square of the rise divided by twice the slope distance. Thus: rise=14 tt:,
<br />slope distance=302.6 ft. Horizontal distance=302.6— 14 X 14 =302.6-0.32=302.28 ft.
<br />2 X 302.6
<br />MADE U4 Y. 11. AL
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