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TRIGONOMETRIC FORMULtE
<br />- B B , B
<br />— ° c a c a c a
<br />A A
<br />b C b _ .0 �, ` C
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />i t a b a b c e
<br />For Angle A. , sin = e ,cos = c tan = b , cot = a , sec = b , cosec =
<br />{�� a ` Z) Given Required a a
<br />�. z
<br />�qa c,vAyl It ' :a,b A,B,e tan A=b= cot B,a a2 { 2=a 1+ a
<br />2
<br />A, B, b sin A = a = cos B, b = (c l a) (tea) = c 1 _ a2
<br />"i c
<br />t --A, a B, b, c B=90°—A, b= acotA,c=
<br />`�G tit! sin A.
<br />{� t 2' �2 a S - �, b B, a; c B = 90°—A, a b tan A, c = b
<br />cos A.
<br />til , 7 ; .7 Z z A, c B, a, b B = 90'—A ix = c sin A, b = c cos A,
<br />X�5_
<br />/
<br />'.o' �-' - Solution of Oblique Triangles
<br />0 6 7- 3 e Given Required
<br />�S % _ „ / . A; B, a b, c, C b = a sin B C = 180°—(A + B), c = a sin C
<br />0 v/ o sin A' sin A
<br />/° 3 b sin A a sin C
<br />0 0 J \ �� Z � 3 0 A.'a, b B,;c C sin B= a ,C' = 180°—(A + B), c = _sin A
<br />(a=b) tan '- (A -f
<br />-
<br />AB)
<br />3 0
<br />-b, C- A, B _ c A+B=180°— C, tan 2 (A—B)= a + b
<br />C7 _ a sin C
<br />1�v. sin A
<br />a+b Fe I(s—
<br />N a, b, e A, B, .0 s = 2 ,sin ;A= be
<br />sin 'B=
<br />2— ac C',—
<br />2 — Y =180° —(A+B)
<br />Z 3-v �.i.a.
<br />1 6 va+b+e
<br />,V fn o r , .c. Area 4= 2 , area =: .s(s—a (s—b (s—c
<br />I A b, c
<br />- � ( Area' b e sin A
<br />� area = 2
<br />y n A 7 /�`
<br />p / I area =
<br />a2 sin B sin C
<br />`j 4' 1 /O "� I A, B, C a Area 2sin A
<br />�i REDUCTION TO HORIZONTAL
<br />i U L �n , - Horizontal distance = Slope distance multiplied by the
<br />V cosine'of the vertical angle. Thus: slope distance =319.4 ft.
<br />r /� I tarts Vert. angle= 50 101. From Table, Page IX. cos 50 10'=
<br />a�0 d 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />f
<br />so pe ngXe a Horizontal distance also =Slope distance minus slope
<br />Q (�. ..A- - distance times (1—cosine of vertical angle). With the
<br />VQ
<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 />r 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 />G �I ante less the square of the rise divided by twice the slope distance. Thus: rise =14 ft..
<br />/ - n slope distance=302.6 ft. Horizontal distance=302.6— 14 X 14 =302.6-0.32=302.28 ft.
<br />\ 'h /, \: 2X302.6
<br />tl sine u+ U. S. a
<br />�O - — -- A
<br />
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