in a cyclic quadrilateral find a,b,c,d . If a=2x-4, b=X+10,c=2y-4 and d=5y+5
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Let ABCD be a cyclic quadrilateral.
∠A=2x+4,∠B=y+3,∠C=2y+10,∠D=4x−5
In cyclic quadrilateral the sum of the opposite angles in 180°. Therefore,
∠A+∠C=180°
⇒2x+4+2y+10=180°
⇒2x+2y=166°
⇒x+y=83°→1
∠B+∠D=180°
⇒y+3+4x−5=180°
⇒4x+y=182°→2
Solving 1 and 2, we get
4x+y−x−y=182°−83°⇒3x=99°⇒x=33°
& 33°+y=83°⇒y=83°−33°=50°
∴∠A=2×33°+4=70°,∠B=50°+3=53°
∠C=2×50°+10=110°,∠D=4×33°−5=127°
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