Suppose Q is a point on the circle with centr | vedclass

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(B) Given that $PQ = 1$ (radius of the circle) and $QR = 1$. In $	riangle PQR$,since $PQ = QR = 1$,it is an isosceles triangle. Therefore,$\angle QPR

Vedclass · Accepted Answer

$87$

Vedclass · Answer

(B) Given that $PQ = 1$ (radius of the circle) and $QR = 1$. In $	riangle PQR$,since $PQ = QR = 1$,it is an isosceles triangle. Therefore,$\angle QPR = \angle QRP = 2^{\circ}$. The sum of angles in $	riangle PQR$ is $180^{\circ}$,so $\angle RQP = 180^{\circ} - (2^{\circ} + 2^{\circ}) = 176^{\circ}$. Now,consider $	riangle S P Q$. Since $SP = SQ = 1$ (radii of the circle),it is an isosceles triangle. Thus,$\angle SQP = \angle QSP$. In $	riangle SPQ$,$\angle SPQ = 2^{\circ}$,so $\angle SQP = \frac{180^{\circ} - 2^{\circ}}{2} = \frac{178^{\circ}}{2} = 89^{\circ}$. Finally,$\angle RQS = \angle RQP - \angle SQP = 176^{\circ} - 89^{\circ} = 87^{\circ}$.

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