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<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin= o , cos- a , tan = b , cot = a , sec = b , cosec -
<br />Given Required
<br />gbA,B,c tan A=b=cotB,a= . as {s=a 41+T
<br />a, G A, B, b sinA = a =cosB b= c a a9
<br />02
<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, e = b
<br />' cos A.
<br />A, c B, a„ b B = 90°—A, a = c sin A, b = c cos A,
<br />Solution of Oblique Triangles
<br />(liven Required a sin B
<br />A, B,a b, c, C b= ,C=180°—(A } B),c=a.sinC ,
<br />BIII A � BlII A
<br />b sin A a sin C
<br />a ti.
<br />A,, a, b B, c, C • sin B = , C = 180°—(A + B) , c = sin A
<br />a,: b, C A, B, c A+B=180°—C,(tan $(A—B)—(a—b) a + bA+B)' +; f
<br />c=
<br />a sin C
<br />sin A
<br />a, b, c A, B, C s=as+,sinA=`I(a .b(c—c
<br />sin #B=1TC=180°—(A+B)
<br />ac
<br />s, b, o Area s— a+b+c2 ,area = 1 a(a—a) (a— s—c
<br />A, b, c Areab e sin A
<br />area = 2
<br />A, B, C, a Area arca = a$ sin B sin C s
<br />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 />a�sI 1c, d Vert. angle =51 M From Table, Page IX. cos 50 10'=
<br />9959. "Horizontal distance=319.4X.9959=318.09 ft. 1i
<br />ca�oPe n�,1e Horizontal distance also =Slope distance minus slope
<br />P a distance times (1—cosine of vertical angle). With the
<br />4e same figures as in the preceding example, the follow.
<br />Horizontal distanceing result is obtained. Cosine 50 101=.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=302:6 ft. Horizontal distance=3026— 14 X 14 =302.6-0.32=302.28 ft.
<br />2 X 3026
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<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin= o , cos- a , tan = b , cot = a , sec = b , cosec -
<br />Given Required
<br />gbA,B,c tan A=b=cotB,a= . as {s=a 41+T
<br />a, G A, B, b sinA = a =cosB b= c a a9
<br />02
<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, e = b
<br />' cos A.
<br />A, c B, a„ b B = 90°—A, a = c sin A, b = c cos A,
<br />Solution of Oblique Triangles
<br />(liven Required a sin B
<br />A, B,a b, c, C b= ,C=180°—(A } B),c=a.sinC ,
<br />BIII A � BlII A
<br />b sin A a sin C
<br />a ti.
<br />A,, a, b B, c, C • sin B = , C = 180°—(A + B) , c = sin A
<br />a,: b, C A, B, c A+B=180°—C,(tan $(A—B)—(a—b) a + bA+B)' +; f
<br />c=
<br />a sin C
<br />sin A
<br />a, b, c A, B, C s=as+,sinA=`I(a .b(c—c
<br />sin #B=1TC=180°—(A+B)
<br />ac
<br />s, b, o Area s— a+b+c2 ,area = 1 a(a—a) (a— s—c
<br />A, b, c Areab e sin A
<br />area = 2
<br />A, B, C, a Area arca = a$ sin B sin C s
<br />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 />a�sI 1c, d Vert. angle =51 M From Table, Page IX. cos 50 10'=
<br />9959. "Horizontal distance=319.4X.9959=318.09 ft. 1i
<br />ca�oPe n�,1e Horizontal distance also =Slope distance minus slope
<br />P a distance times (1—cosine of vertical angle). With the
<br />4e same figures as in the preceding example, the follow.
<br />Horizontal distanceing result is obtained. Cosine 50 101=.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=302:6 ft. Horizontal distance=3026— 14 X 14 =302.6-0.32=302.28 ft.
<br />2 X 3026
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