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<br /> Aj
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<br /> 53 l, I1
<br /> a
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<br /> /1 ' 4 C s!. �G ? Q 1 b O b
<br /> V., g +, Right Tribe L�Obligee Trio'ag1j,
<br /> 1 '/. 40 Solution of
<br /> Right TntsEnglea.
<br /> 9 7
<br /> _-�. s- Poe AnoeA� tin.s 0,--C,tan– b,cOi_ a C 0
<br /> • i 2 7� .�. / 3 - _ 6, b r cosec m
<br /> Given- Required a
<br /> /`/ a, b A,B,c tan,t='b a Cot$,a N/aT+ a V1+
<br /> a'
<br /> FJ`_3 a, c A B, b naA=a—008B.b—%I +a c—a sa 1—a
<br /> d.7
<br /> A,a '$r b, a B=90°—A,b =acot A,6a a
<br /> i! sin A.
<br /> Y�i 5 t i y� 9, b B,a, c B-900—A,a = b tan A,c=
<br /> 7 J cos A. j �_3
<br /> A, o B, a, 'b B—90°A,a =e sin A,b= e cos A,
<br /> 974 37•4dri"ti/ Solution of Oblique Triangles 3 1
<br /> 1/ (liven Required a sin B a sin s
<br /> , d, B,a b, c, C h= ' C= 180o—(A+B),
<br /> vsinAc =
<br /> !' sin A
<br /> b sin A
<br /> � s!, a, b B, c, C ein B= a ,C= 180°—(A-r-B),o = a sia a
<br /> sine
<br /> a, b, O A, B,c A+B=180°—C,tan,}(A-.-B)—(a—b) jn')(A+'B)
<br /> asin a a+ b
<br /> sin A
<br /> 3. 8�' ` a, b, o A B, C a=a+— b+
<br /> 2 ,ein}A=
<br /> sin B=
<br /> 2 "- a c 'C=180°—(A+B)
<br /> a, b, a Area 8= +b c, area = s(s—a e—
<br /> A, b, c Area area = bas^ in L l J
<br /> as sin B sin a
<br /> Area area _ ------- 7
<br /> U d d D REDUCTION TO HORIZONTAL
<br /> Horizontal dis hm'&Slope distance aalt[pW b7 the
<br /> VerL ofthe vertical#Psa1e.•4hus.slope
<br /> r a� 1e=6'•10.- From T,h1%ped I7L O'b id
<br /> h- .�•' 3 eYi® xerimnta3'dLtanoe—M&W.ai9=3"tL
<br /> 1'O Hod rizo talmeistance also eta diehnee minus slop
<br /> same fhmree as is the p • coWith file
<br /> 'l inQ result is oi =
<br /> AWL
<br /> OW-
<br /> - 818 4X.00II=.1.91.E1&#-181=fif108 it
<br /> iV3ea th(triae is known,the herizotital distance is aDproslaaatel .—the slope dist-
<br /> -
<br /> �' sly�tbw sgture of the rise divided by tadoe theX�k �va:rile=edit..
<br /> pe distaaoe-8Q8•# .
<br /> _ '` . - a�c.am e'•g°�a-off_ ffi it.
<br /> NAN IN Y.N.A.
<br /> e
<br /> IN
<br /> e
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