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I
<br />TRIGONOMETRIC FORMULA �
<br />i� B B B
<br />c ri a c� d c a
<br />A b £CA b XCA b AC
<br />Right Triangle Oblique Triangles !J
<br />Solution of Right Triangles
<br />b a b c e
<br />For Angle A. sin = a , cos = o , tan= b , cot = a , sec = b , cosec = a
<br />Given Required a 2
<br />a,b A,B,c tan A=b=cot B,c= a2+ 2=a 1+ d2
<br />a, -c A, B, b sin A = o = cos B, b — %.c --
<br />/ — a (ca = c
<br />A, a B, b, a B=90°—A, b= a cotA, c= . a
<br />sin A.
<br />A, b B, a, c B=90°—A,a _ btanA,c= b
<br />cos A.
<br />A, c B, a, b B = 900—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, C b = sin A , C = 180°—(A + B), c = sin A
<br />b sin Aa sin C
<br />A, a, b B, c, O sin B = a , C = 180°—(A + B) , c = sin A
<br />a, b, C A, B, c A+] =180°— C, tan i (A—B)= (a—b) tan 'j' (A+B)
<br />=
<br />a sin C
<br />o
<br />sin A
<br />a, b, o A, B, C .s=a+b+0 .,in 11A= -J bG—c .
<br />1i sinB= a o_ 10=180°—(A+B)
<br />a, b, c Areas=a+2+c,
<br />A; b, c Area b c sin A
<br />area = 2
<br />a2 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 />_ gree Vert. angle =50 101. From Table, Page IX. cos 50 lot=
<br />e 6' e v 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />SN n��e Horizontal distance also=Slone distance minus slope
<br />Ve . A a 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 51 101=.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 />ance less the square of the rise divided by twice the slope distance. Thus: rise=14 ft.,
<br />slope distance=3026 ft. Horizontal distance=3026— 14 X ]4 =302.6-0.32=302 28 ft.
<br />2 X 302.6
<br />MADE U1 Y. S. A.
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