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Sum of Opposite Angles of a Cyclic Quadrilateral

In a cyclic-q...

Question

In a cyclic-quadrilateral PQRS, angle $P Q R = 135^{\circ} .$ Sides SP and RQ produced meet at point A : whereas sides PQ and SR produced meet at point B.

If $∠ A : ∠ B = 2 : 1;$ find angles A and B.

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Solution

Given $PQRS$ is a cyclic quadrilateral.

$\angle A = \angle B = 2 :1$ or if $\angle B = x$ , then $\angle A = 2x$

$\angle PQR +\angle PSR = 180^o$ (sum of the opposite angles in a cyclic quadrilateral is $180^o$ )

$\angle PSR = 180-135 = 45^o$

In $\triangle PBS; \angle SPB + 45+x = 180^o$

$\Rightarrow \angle SPB = 135-x$ … … … (i)

In $\triangle PQA; \angle APQ = 180-2x-45 = 135-2x$

Therefore $\angle SPB = 180 - (135-2x) = 45+2x$ … … … (ii)

Hence $35-x = 45+2x \Rightarrow x = 30^o$

Therefore $\angle B = 30^o \ and \ \angle A = 60^o$

$\\$

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