In a circle with centre O,PQ and XY are chord | vedclass

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(D) In $	riangle OXY$,$OX = OY$ (radii of the same circle).Therefore,$\angle OYX = \angle OXY = 30^{\circ}$.In $	riangle OXY$,the sum of angles is $

Vedclass · Accepted Answer

$8$

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(D) In $	riangle OXY$,$OX = OY$ (radii of the same circle).Therefore,$\angle OYX = \angle OXY = 30^{\circ}$.In $	riangle OXY$,the sum of angles is $180^{\circ}$,so $\angle XOY = 180^{\circ} - (30^{\circ} + 30^{\circ}) = 120^{\circ}$.Using the Law of Sines in $	riangle OXY$: $\frac{XY}{\sin(120^{\circ})} = \frac{OX}{\sin(30^{\circ})}$.$OX = \frac{8 \cdot \sin(30^{\circ})}{\sin(120^{\circ})} = \frac{8 \cdot 0.5}{\sqrt{3}/2} = \frac{8}{\sqrt{3}}$.Now,in $	riangle OPQ$,$OP = OQ = OX = \frac{8}{\sqrt{3}}$.Using the Law of Sines in $	riangle OPQ$: $PQ = 2 \cdot OP \cdot \sin(\frac{120^{\circ}}{2}) = 2 \cdot \frac{8}{\sqrt{3}} \cdot \sin(60^{\circ})$.$PQ = 2 \cdot \frac{8}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = 8 \, cm$.

In a circle with centre $O$,$PQ$ and $XY$ are chords. If $\angle POQ = 120^{\circ}$,$\angle OXY = 30^{\circ}$ and $XY = 8 \, cm$,then $PQ$ is ........... $cm$.

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