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(,( `� � TRIGONOMETRIC -FORMIJUE
<br /> vB lq,15" B
<br /> c
<br /> o,
<br /> I Z J 7a a
<br /> _ A A C
<br /> C p C b
<br /> Right` angle Oblique Triangle@
<br /> Solution of Right Triangles
<br /> 7 j For Angle , sin =aa b tan= a b a m
<br /> a , a , T,cot= a,sec—b, co@ec a
<br /> 2 Given 44gaired a
<br /> L 3 v a, b , ,.B,o tan A=b= cot B,a = a$+—F2
<br /> a+ a $, b sin A=a=ew B,b
<br /> �C f A,a s B=90°—A, b =a cotA,c= Bin A.
<br /> 3 1 1 0°et;b•, fit, a B=90°—A,a = b tan A,a= cos A.
<br /> D A,a B, a, b B=90°—A,a=a sin A,b=o cos A,
<br /> Solution of Oblique Triangles
<br /> tlivea'. .Sequined Solution
<br /> sin B
<br /> �Gl ) j A. B.
<br /> �, a; a, C b= Bin A ' C= 180°=(A ., �.a = Bin A
<br /> A, a, b B,c, C Bin B= b sin AC= 180*—(A t-B),c = a Bin C
<br /> a sin A
<br /> a, b,17 'd,B,a A+B_1800—C,tan I(a BB)C a—b t+i(bA-FB)
<br /> in
<br /> B1n AA .R.
<br /> - _
<br /> 4.b, 0rn8, C s=a Zsin)A=
<br /> i,
<br /> ein}B=�" 40 ,C.180°—(A+B)
<br /> @=a+b+a� area
<br /> J G A, b, a y- b a Bin A
<br /> rl- area =
<br /> A B C a ar sin B sin C
<br /> , AtGa area = A
<br /> rj c REDUCTION TO HORIZONTAL,
<br /> J JioBizoaatt��l distance—Slope distance
<br /> .mnitlplied by the
<br /> ooe a of the veGd �Thns:slope distance=f18 4 ft.
<br /> \ \ angle=b m Tabl page X.cos b°10+=
<br /> } liorizoa distances=819 4X. 8 00 ft,
<br /> �l • '4 ~ \ s` {. .` Looe f•�le V orizonhl dislaac®also—Slave disYu►oe minas slope
<br /> U\ �►\ Lex '� Q distance times fl-cosine of vertical angW With the
<br /> tt . same fiQ,tres,as in the preceding example,the follosr-
<br /> 7 Horizontal distance ing resat is obtained.Cosine b°101=.99%i�."pD=A 41•
<br /> 81
<br /> When the rise is known, Y.4X 0041=1.51.819.4-1.91=818.00 ft.
<br /> own,the horizontal distanoe fs approxlmately:—the elope idlW
<br /> f snee less the square of the rise divided•by twioe the slope distance. Thust rlse�asll fi;
<br /> -slope distance-one ft. Horizontal 14 X =gg•g.._0M—M2g
<br /> 2 X 802.6
<br /> VIA ere►e.A. ;
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