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2015-03-09T00:17:30+05:30

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As it is a cyclic quadrilateral../_A+/_C=/_B+/_D=180......2x+4+2y+10=180.....2x+2y+14=180...,,x+y+7=90.....x+y=83..eq1.....Now for eq2..y+3+4x-5=180....4x+y-2=180...4x+y=182..y=182-4x NOW SUBSTITUTING THE VALUE OF y..in eq1 we have..x+182-4x=83...-3x=-99...x=33....Again in eq1..x+y=83....33+y=83....y=50..(Ans) /_A=2*33+4=70.../_B=y+3=53.../_C=2y+10=110.../_D=4x-5...132-5=127
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2015-03-09T00:22:24+05:30

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/_ A = (2x+4)°, /_B = (y+3)° , /_ C= (2y+10)°,/_D=(4x-5)°

Now, /_A+ /_C = 180° ( opp angles of a cyclic quad are supplementary ).......1

and /_B+/_D = 180 ( same as above)................2

putting the values in 1 and 2,(2x+4)° +  (2y+10)° = 180°

 Slly, (y+3)° +(4x-5)°= 180°.= (4x + y - 2)= 180. 

multiplying eq. 1 by 2, 2 X {(2x+4)° +  (2y+10)° = 180°}

 = 2X ( 2x+2y+14)=4x + 4y +28.=360°

Finally, subtracting 2 from 1

,(4x + 4y +28.=360°) -  (4x + y - 2)=180.

= 3y + 30 = 360-180= 180.

3y = 180-30=150.

 y = 150/3 = 50°.

 putting value of y in eq 2.

(4x + 50 - 2)= 180.

we get x = 180-48/4 = 33°.

Hence, /_A= 2x33+4= 70°/_ B = 50+3= 53°/_C= 180 - 70 = 110°/_D= 180-53 = 127°

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dat was a long ans...took a while to complete :0
Its ok^^ & thanks a lot a lot a lot