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Casio. Cotg. Case& Sec. Tan. Sin. Angle
<br />it I,
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<br />Casio. Cotg. Cosec. Sec. Tan. Sin. (a-7))) tan Angl
<br />TRIGONOMETRIC FORMULA
<br />B B
<br />a
<br />c a c a c
<br />s J� 3
<br />`� b CAA b C A b C l
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />y a b a_
<br />For Angle A. sin = — , cos = — , tan = cot = — a , sec = —, cosec =
<br />c c b
<br />Given Required a
<br />a,b I A,B,c taaA=b= cot B,c= a" —+-7 _=a, i+ x2
<br />a, c A, B, b 0.2
<br />A, a B, b, c B = 90°—A, b = a cot A, e, = a -17
<br />1 �.
<br />sin _4.
<br />A, b B, a, c B = 90°—A, a = b tan A, c = cos A. ' tt'
<br />�) <'
<br />A, c If, a., b B = 90'--4, a = o sin A, b = 6 COS A,: - U
<br />Solution of Oblique Triangles 3
<br />Given Required a sin B a sin C,1
<br />A, B,a b, c, C b= C=180°—(A B),c=
<br />sin A sin A
<br />b sin A a sin C t
<br />A, a, b 11, c, C sin B= G' a C= 180°—(1 -( B), c = sin A
<br />(
<br />a, b, A, B, e A+B=180°— G', ten :': (A—B)_ ('- A+B
<br />�,
<br />a + b
<br />c — a sin C y�rI
<br />sin A ? D
<br />a, b, c A, B, C 4= 2 sin] ._4=, be
<br />f '7
<br />sin B=1 C=180°,=(A+B)
<br />ac
<br />a, b e Area s—«+�+c, area = a(a—a s— ){,—c)
<br />A, b, c Area b c sin A
<br />area
<br />= a2 sin B sin C
<br />2
<br />A, B, G; a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />e Horizontal distance= Slope distance multiplied by the
<br />cc cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />6""", Vert- angle =5° 30'. From Table, Page IX. cos 50 101=
<br />e H 9959. Horizontal distance=310.4X.9959=315.09 ft.
<br />$�°� A�'t" Horizontal distance also= Slope distance minus Slope
<br />�e 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 10'=.9959. 1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=31&09 ft.
<br />When the rise is known, the horizontal distance is approximately:—tbe 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 =302.6-0.3 =30225 it
<br />2 X 302 6
<br />MADE IN U.S.A.
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<br />Casio. Cotg. Case& Sec. Tan. Sin. Angle
<br />it I,
<br />13
<br />Casio. Cotg. Cosec. Sec. Tan. Sin. (a-7))) tan Angl
<br />TRIGONOMETRIC FORMULA
<br />B B
<br />a
<br />c a c a c
<br />s J� 3
<br />`� b CAA b C A b C l
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />y a b a_
<br />For Angle A. sin = — , cos = — , tan = cot = — a , sec = —, cosec =
<br />c c b
<br />Given Required a
<br />a,b I A,B,c taaA=b= cot B,c= a" —+-7 _=a, i+ x2
<br />a, c A, B, b 0.2
<br />A, a B, b, c B = 90°—A, b = a cot A, e, = a -17
<br />1 �.
<br />sin _4.
<br />A, b B, a, c B = 90°—A, a = b tan A, c = cos A. ' tt'
<br />�) <'
<br />A, c If, a., b B = 90'--4, a = o sin A, b = 6 COS A,: - U
<br />Solution of Oblique Triangles 3
<br />Given Required a sin B a sin C,1
<br />A, B,a b, c, C b= C=180°—(A B),c=
<br />sin A sin A
<br />b sin A a sin C t
<br />A, a, b 11, c, C sin B= G' a C= 180°—(1 -( B), c = sin A
<br />(
<br />a, b, A, B, e A+B=180°— G', ten :': (A—B)_ ('- A+B
<br />�,
<br />a + b
<br />c — a sin C y�rI
<br />sin A ? D
<br />a, b, c A, B, C 4= 2 sin] ._4=, be
<br />f '7
<br />sin B=1 C=180°,=(A+B)
<br />ac
<br />a, b e Area s—«+�+c, area = a(a—a s— ){,—c)
<br />A, b, c Area b c sin A
<br />area
<br />= a2 sin B sin C
<br />2
<br />A, B, G; a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />e Horizontal distance= Slope distance multiplied by the
<br />cc cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />6""", Vert- angle =5° 30'. From Table, Page IX. cos 50 101=
<br />e H 9959. Horizontal distance=310.4X.9959=315.09 ft.
<br />$�°� A�'t" Horizontal distance also= Slope distance minus Slope
<br />�e 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 10'=.9959. 1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=31&09 ft.
<br />When the rise is known, the horizontal distance is approximately:—tbe 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 =302.6-0.3 =30225 it
<br />2 X 302 6
<br />MADE IN U.S.A.
<br />
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