|
B
<br />TRIGONOMETRIC FORMULAE I
<br />3, B B
<br />c a ° a c a
<br />5
<br />A b C A�b C d C
<br />Right Triangle Oblique Triangles ��
<br />Solution of Right 'Triangles
<br />For Angle A. sin = a , cos = c b , tan= a , cot = b a , sec = b a , cosec = o
<br />c b a
<br />Given Requireda
<br />a, b A, B ,c tan A = b =cot B, c = a2 -+b2 = a 1 + az
<br />a, c A, B, b 1_2L s
<br />A, a B, b, e B=90°—A, b = a cot A, c= a -
<br />sin A.
<br />A, b B, a, c B = 90°—A, a = b tan A, c = b
<br />cos A,
<br />A, r. B, a, b I B = 90°—A, a = c sin A, b = c cos A,
<br />Solution of Oblique Triangles
<br />Given Required _ a sin B
<br />A, B, a. b, c, b sin A , C = 180°—(A + B), c = sin A
<br />A, a, b ' B, c, C sin B = a b sin A, C = 180°—(A { B), c = a sin C
<br />sin A
<br />1 (a—b) tan z (A+:q)
<br />- a, b, C A, B, c - A-}-8=180 � C, tan 2 (A—B)= a + b ,
<br />c=
<br />a sinC
<br />sin A
<br />a, b, c A, B, C s=a+2+c,,in aA= V (s--b(c—c r
<br />�—c
<br />sinzB= (4—a)(S ,C=180°—(A+B)
<br />ac
<br />a, b, c Area s=a+2+c,
<br />A, b, c Area area = b e sin A
<br />2
<br />az sin B sin C
<br />A, B, C, a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />farce Vert. angle =5° 101. From Table, Page IX. cos 5° 10-
<br />e aj5 9959. Horizontal distance -319.4X.9959=318.09 ft.
<br />Horizoslope t. An,c C4 distance timesal (1 distance cosine oflvertical a angle)nce . With niis lthe
<br />v same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5° 10'=.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 />ance less the square of the rise divided by twice the slope distance. Thus: rise=14 ft.,
<br />slope distance=302.6 ft. Horizontal distance=302.6— 14 X 14 =392 6-0.32=302.28 ft
<br />2 X 302.6
<br />MADE W u. s. A.
<br />
|