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<br />I92 L� TRIGONOMETRIC FORMULiR
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<br />0� jil 9 p!� v p 2.L = \. A� B a b' �' C b sin nA ' C = 180°—(A } B), c =a sin s A
<br />b sin A a sin C
<br />A, a, b B, c, C sinB= ,C= 180°—(A+B),c =
<br />� B B
<br />.
<br />a { b
<br />a sin C
<br />c
<br />a c a c a
<br />.e.'
<br />C
<br />A�b C A � C
<br />b,
<br />'
<br />Right Triangle
<br />Oblique Triangles
<br />A, b, c Area .area = bcsin A
<br />2
<br />Solution of Right Triangles -
<br />)
<br />For Angle A, sin = � ,cos = � ,tan = b cot=—,sec= b , cosec = a
<br />�
<br />Given
<br />a,b
<br />Required
<br />A,B,c
<br />tanA=b=cotB,c= as-}- a=a 1+ as
<br />�
<br />a, e
<br />'A, B, b
<br />a
<br />sin'A = � =cos B, b = �/ (c+a) (c—a} = c � 1— a �
<br />tar°e Vert. angle =5° 101. From Table, Page IX. cos 5° 101=
<br />e ass y 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />A', a
<br />B, b, c
<br />B=90°—A, b= a cotA, c= a
<br />Vet. distance times g—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />sin A.
<br />Horizontal distance ing result is obtained. Cosine 50 101=.9959.1—.9959=.0041.
<br />A, b
<br />B, a, c
<br />B=90'—A, a = b tan A, c = b
<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 />cos A.
<br />=302. ' :slope distance=302.0 ft. Horizantal'distance=30214 X 14 6— 8-0.32=302.28 ft.
<br />A, c
<br />B, a, b
<br />B = 90°—A, a = c sin A, b = c cos A,
<br />MADE IM U. B. A.
<br />Solution of Oblique- Triangles
<br />Given
<br />Required
<br />B C
<br />A, a, b B, c, C sinB= ,C= 180°—(A+B),c =
<br />a sin A
<br />a, b, . C A, B, c A+B=1800— C, tan 3 (A—B)= (a—b) tan ; (A+B)�
<br />a { b
<br />a sin C
<br />c = sin A
<br />a+b+c
<br />�(s—
<br />a, b, c A, B, C s= 2 ,sin -',A= b c '
<br />sin 3.B- J ,C=180°—(A+B)
<br />a c
<br />}a,+b+c
<br />r
<br />! a, b, c Area s = 2 ,area (s—e
<br />A, b, c Area .area = bcsin A
<br />2
<br />as sin B sin C
<br />A, B, C, a Area area =
<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 />!
<br />tar°e Vert. angle =5° 101. From Table, Page IX. cos 5° 101=
<br />e ass y 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />gNoQ AogNe a Horizontal distance also=Slope distance minus slope
<br />Vet. distance times g—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 50 101=.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 />=302. ' :slope distance=302.0 ft. Horizantal'distance=30214 X 14 6— 8-0.32=302.28 ft.
<br />° 2 X 302.0
<br />MADE IM U. B. A.
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