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►= y �^� � moo. -a �
<br />O `fi tr. "-
<br />1
<br />o ° -3- Z
<br />313, '�s
<br />r
<br />1 2 be
<br />Yin, 09
<br />7210)'
<br />�f
<br />siri;B= ),C=180°—(A t B)
<br />r a+b+c
<br />*.b, c Area s= 2 , area
<br />area =
<br />A, b, e, 'Area b c sin A
<br />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 />Vert. angle =5° 1tN. From Table, Page IX. cos 51101
<br />ass
<br />49959. Horizontal distance=319.4X.9959=318.09 ft.
<br />Horizontal distance also = Slone distance minus slope
<br />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 50 lol=.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 =302.6-0.32=3M28 ft.
<br />____2X302.6
<br />MADE In.U. a.N
<br />11�
<br />TRIGONOMETRIC FORMULAE
<br />B B B
<br />-I
<br />a
<br />a c a c a
<br />d b
<br />Right Triangle
<br />C A� b C A b C
<br />`-- Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle
<br />A. sin =
<br />a b a b c a%
<br />c. , cos = c , tan= b , cot = a ,'sec -F b , cosec
<br />Given '
<br />a, b
<br />Required
<br />A, B ,c
<br />z
<br />tan A = b =• cot B, c = a2 +.
<br />I ? ta, a
<br />d, B, b
<br />2
<br />sin A = c = cos B, b = �/ (c+a (e—a) = c 1— c
<br />i. A, a
<br />B, b, c
<br />B=90°—A, b = a cotA, c=
<br />sin A.
<br />b
<br />'A, b
<br />B, a, c
<br />- B = 90°—A, a = b tan A, c =
<br />cos A. .
<br />A, c
<br />B, a, b
<br />B =900—A, a = c sin A, b = c cos A,
<br />Solution of Oblique Triangles
<br />x1 Given
<br />A, B
<br />Required
<br />b, c, C
<br />a sin B a sin C
<br />b= --—(AB),c= sin
<br />sinA'C=180,
<br />\\ 1
<br />C
<br />b sin A a sin C
<br />sin B = , C = 180 —(A 1 B) e =
<br />a sin A
<br />a,. b, C
<br />A, B, c
<br />(a—b) tan z (A+B)
<br />A+B=180o — C, tan (A—B)=
<br />a + b
<br />a sin C
<br />-
<br />_
<br />e sin A'
<br />?� a, b a
<br />'A B, -C
<br />a+b+c
<br />s=—,sin gA-- ,
<br />1 2 be
<br />Yin, 09
<br />7210)'
<br />�f
<br />siri;B= ),C=180°—(A t B)
<br />r a+b+c
<br />*.b, c Area s= 2 , area
<br />area =
<br />A, b, e, 'Area b c sin A
<br />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 />Vert. angle =5° 1tN. From Table, Page IX. cos 51101
<br />ass
<br />49959. Horizontal distance=319.4X.9959=318.09 ft.
<br />Horizontal distance also = Slone distance minus slope
<br />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 50 lol=.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 =302.6-0.32=3M28 ft.
<br />____2X302.6
<br />MADE In.U. a.N
<br />11�
<br />
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