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<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin = a , cos = o , tan = b , cot = b, sec =. b , cosec = a
<br />Given Required
<br />a, b A, B ,c tan A = b = cot B, a = a2 -} 2 = a 1 a2
<br />a, c A, B, b sin A = m = cos B, b = c a c—a az
<br />03
<br />A, a B, b, c B=90°—A, b= a cotA, 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, c . 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 in C
<br />A, B, a b, c, C b sin A , C = 180°—(A + B), c = sin
<br />A, a, b B, c, C sin B = b sin A , C = 1800—(A { B) a sin C
<br />a , c = sin A
<br />a, b, C A, B, c A+B-180°— C, ten 3 (ab)tan 1
<br />(A—B)= a �' (A+B)
<br />b ,
<br />a sin C
<br />sin A
<br />a, b, c A B C s-a+2+c,sin;A=,I(s— bc—cam
<br />(,q—a)(,9)'
<br />sin,B= C=180°—(A } B)
<br />ac
<br />a, b, c Area s = a+2
<br />A, b, c Areab c sin A
<br />area = 2
<br />as 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 />e a�s�a�ce y 9959. Horizontal distancem 19.4X 959= 318.09 ft Table, PaIX. cos 5° 10
<br />$1ov "e a Horizontal distance also= Slope distance minus slope
<br />Ve 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.h'
<br />When the rise is known, the orizontal 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 =3026-0.32=302.28 M
<br />2 X 302.6
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<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin = a , cos = o , tan = b , cot = b, sec =. b , cosec = a
<br />Given Required
<br />a, b A, B ,c tan A = b = cot B, a = a2 -} 2 = a 1 a2
<br />a, c A, B, b sin A = m = cos B, b = c a c—a az
<br />03
<br />A, a B, b, c B=90°—A, b= a cotA, 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, c . 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 in C
<br />A, B, a b, c, C b sin A , C = 180°—(A + B), c = sin
<br />A, a, b B, c, C sin B = b sin A , C = 1800—(A { B) a sin C
<br />a , c = sin A
<br />a, b, C A, B, c A+B-180°— C, ten 3 (ab)tan 1
<br />(A—B)= a �' (A+B)
<br />b ,
<br />a sin C
<br />sin A
<br />a, b, c A B C s-a+2+c,sin;A=,I(s— bc—cam
<br />(,q—a)(,9)'
<br />sin,B= C=180°—(A } B)
<br />ac
<br />a, b, c Area s = a+2
<br />A, b, c Areab c sin A
<br />area = 2
<br />as 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 />e a�s�a�ce y 9959. Horizontal distancem 19.4X 959= 318.09 ft Table, PaIX. cos 5° 10
<br />$1ov "e a Horizontal distance also= Slope distance minus slope
<br />Ve 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.h'
<br />When the rise is known, the orizontal 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 =3026-0.32=302.28 M
<br />2 X 302.6
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