The triangle PQR is inscribed in the circle x | vedclass

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(C) The given equation of the circle is $x^2+y^2=25$. Therefore,the centre $O$ is $(0,0)$ and the radius $r=5$. Now,calculate the distance $QR$: $QR =

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$\frac{\pi}{4}$

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(C) The given equation of the circle is $x^2+y^2=25$. Therefore,the centre $O$ is $(0,0)$ and the radius $r=5$. Now,calculate the distance $QR$: $QR = \sqrt{(-4-3)^2 + (3-4)^2} = \sqrt{(-7)^2 + (-1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$. Since $O$ is the centre,$OQ = OR = 5$. Using the law of cosines in $	riangle OQR$: $\cos(\angle QOR) = \frac{OQ^2 + OR^2 - QR^2}{2 	imes OQ 	imes OR} = \frac{25 + 25 - 50}{2 	imes 5 	imes 5} = \frac{0}{50} = 0$. Thus,$\angle QOR = \frac{\pi}{2}$. We know that the angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the remaining part of the circle. Therefore,$\angle QOR = 2 \angle QPR$. $\frac{\pi}{2} = 2 \angle QPR \implies \angle QPR = \frac{\pi}{4}$.

The triangle $PQR$ is inscribed in the circle $x^2+y^2=25$. If $Q=(3,4)$ and $R=(-4,3)$,then $\angle QPR=$

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