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In a cyclic quadrilateral `A B C D` , `/_A=(2x+4)o, /_B=(y+3)o, /_C=(2y+10)o, /_D=(4x-5)o` . Find the four angles.

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Correct Answer - `angle A = 70^(@), angle B = 53^(@), angle C = 110^(@), angle D = 127^(@)`.
We have , `angle A + angle C = 180^(@) and angle B + angle D = 180^(@)`.
` therefore ( 2 x + 4) + ( 2y + 10 ) = 180rArr x + y = 83" " `… (i)
and ` ( y + 3 ) + ( 4x - 5) = 180rArr 4x + y = 182 " " `... (ii)
From (i) and (ii), we get
` therefore 3x = 99rArr x = 33. And, 33 + y =83 rArr y = 50`.

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