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-2,- 3-1-
<br />!z>; 9D �2
<br />�_) Z
<br />z
<br />_ � �A%1a4L11'•�3
<br />53
<br />TRIGONOMETRIC FORMUL1E
<br />B B B
<br />a c a c a
<br />}` b C A b -G'A b C
<br />' `Right:Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. ,sin = a ,cos = ,tan = b ,cot = a ,sec = b , cosec = a
<br />S Given Required
<br />a,b 4,B,c tanA=6= Cot B,c= _+ =a 1+a'
<br />a, o 21, B, b sin A. = a = cos B, b (c=, a) (c—a) = c 1 —
<br />Q$ a '
<br />c 1.
<br />A, a-
<br />B, b, cB
<br />(11
<br />= 90°-A, b = a cotA, e = sin A. --�
<br />� A, b
<br />I`} -7 n
<br />B=90' -A a= btan A,c= b "
<br />cos A..
<br />A, c
<br />-B, a,, b
<br />B=90'—A, a = c sin A, b = e cos A,
<br />Solution of Oblique Triangles
<br />Given
<br />Ct
<br />a sia s C
<br />-2,- 3-1-
<br />!z>; 9D �2
<br />�_) Z
<br />z
<br />_ � �A%1a4L11'•�3
<br />53
<br />TRIGONOMETRIC FORMUL1E
<br />B B B
<br />a c a c a
<br />}` b C A b -G'A b C
<br />' `Right:Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. ,sin = a ,cos = ,tan = b ,cot = a ,sec = b , cosec = a
<br />S Given Required
<br />a,b 4,B,c tanA=6= Cot B,c= _+ =a 1+a'
<br />a, o 21, B, b sin A. = a = cos B, b (c=, a) (c—a) = c 1 —
<br />Q$ a '
<br />c 1.
<br />A, a-
<br />B, b, cB
<br />(11
<br />= 90°-A, b = a cotA, e = sin A. --�
<br />� A, b
<br />B; a, c
<br />B=90' -A a= btan A,c= b "
<br />cos A..
<br />A, c
<br />-B, a,, b
<br />B=90'—A, a = c sin A, b = e cos A,
<br />Solution of Oblique Triangles
<br />Given
<br />Required
<br />a sia s C
<br />A, B, a
<br />b, c, C
<br />b - , C = 180°—(A + B), c = �n
<br />nnA
<br />A, 2, b
<br />B, c, C
<br />b sin A. a sin C
<br />sin B = , C = 180°—(A + B), c =
<br />a sin A
<br />a, b, G
<br />A, B, c
<br />o I (a—b) tan z (A { B)
<br />• A-�=, B=180 — C, tan ,' (A --B)= ,
<br />a + b
<br />a sin C
<br />_
<br />c sin A.
<br />k
<br />a, b, e
<br />A, B, C
<br />s = 2 ,sin 1.4—
<br />be '
<br />C=1801—(A+B) sin'B—�
<br />ac
<br />c,
<br />4=z+�+c, area = �/s(s—a(8—b)(s—c)
<br />.)Arca
<br />)1 A, b, c
<br />Arca
<br />besin A
<br />arca = •
<br />2
<br />a2 sin B sin C
<br />(, A, B, C, a
<br />'Area
<br />area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />"
<br />Horizontal distance=Slope distance multiplied by the
<br />e
<br />cosiuc of the vertical angle. Thus: slope distance=319.4ft.
<br />Vert. angle=5' 101. From Table, Page IX. cos 5'101=
<br />•zontalldistancesalsoe
<br />��o0eV'Xe
<br />t.
<br />x Ho Slone distance8ninus slope
<br />distance times (1—cosine of vertical angle). With the
<br />�e
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance
<br />ingresult is obtained. Cosine 51101=.9959.1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=31309 ft.
<br />n' When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the
<br />rise divided by twice the slope distance. Thus: rise =14 fL,
<br />slope distance=302.6 ft.
<br />Horizontal distance=302.6— 14 X 14 =302.6-0.32=302.28 fL
<br />2 X 302.6
<br />.�'
<br />MADE IN V.10, A.
<br />y
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