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TRIGONOMETRIC FORMUL,-E
<br />,.
<br />cl c f tj :2
<br />to
<br />A C
<br />0:3
<br />CAA b C b
<br />Right Triangle Oblique Triangles J
<br />Solution of Right Triangles
<br />5
<br />For Angle A., sin = ,cos = b ,tan = b , cot = a , sec = 6 , cosec = a
<br />,
<br />Given
<br />a, b
<br />Required
<br />B
<br />az
<br />tan A = = cot B, c = , - = = a 1 f
<br />G 3
<br />.4, ,.c
<br />b az
<br />' ��,"� 1•, 21 tAS
<br />ll a,c
<br />4>B, b
<br />= a = — — — aE n
<br />sin cos B,b—�(DIa)(c a)—e `I1 -0s
<br />�1b
<br />a
<br />c86 ave) 3�"I 4,S
<br />A, a
<br />B, b, c
<br />B=90°—A, b = acotA,c= sin A.
<br />_IrF�^ —}i
<br />A, b.
<br />B, a, c
<br />B = 90'—A, a = b tan A, c =
<br />cos A.
<br />iUt$, G- �?Y fl
<br />A,c
<br />'B,a, b
<br />B=90°—A,(L=esin A,b=ccosA, t+
<br />✓ •�`_�„------ �"
<br />Solution of Oblique Triangles
<br />r) ' I
<br />Given
<br />A, B, (L
<br />Required
<br />b, c, C
<br />a sin B 2sin C
<br />b = C = 180° —(A + B), c =
<br />sin A sin A
<br />C. ob
<br />A,a,b
<br />B,c,C
<br />sin A °— ea sin C
<br />s:nB= ,C=180 (AtB),c=
<br />a sin
<br />°— (a—&)tan(41B)
<br />b
<br />a, C
<br />,
<br />A, B c
<br />A B—
<br />�- C, tan _ 1 (-1—B)= .a -i-
<br />1 1 -t% So' 1 ?j i!'�
<br />_asinC
<br />c
<br />A
<br />- _
<br />sin
<br />a, b, 'c
<br />-9,B,G
<br />a+b-f
<br />s= 2 sini:i= J be
<br />3
<br />s:n:_,B—�I
<br />a c ,C=180 —(A+B)
<br />&
<br />!( a, b, c
<br />Area
<br />s = 2 area = 6(8—a) (s— ) (s—c)
<br />A , b, a
<br />Area
<br />area = b c sin A
<br />2
<br />a"- sin B sin C
<br />-
<br />A, B, Q,a
<br />Area
<br />area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance = Slope distance multiplied by the
<br />-
<br />cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />Vert. angle=50101. From Table, Page IX. cos 5° lo/-
<br />5 e d ,S`o"I
<br />X04 Ze
<br />9959. Horizontal distance=379.4X.9959=315.09 ft.
<br />Horizontal distance also=Slone distance minus slope
<br />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 5°101=.9959. 1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=316.09 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=302.26 fL
<br />-
<br />2 X 302.6
<br />'
<br />MADE IR w a, A.
<br />-
<br />/
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