In a triangle PQR,let angle PQR = 30^{circ} a | vedclass

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(B) Given: $PQ = 10\sqrt{3}$,$QR = 10$,and $\angle PQR = 30^{\circ}$.Using the Law of Cosines to find $PR$:$PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(30^{\ci

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$B, C, D$

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(B) Given: $PQ = 10\sqrt{3}$,$QR = 10$,and $\angle PQR = 30^{\circ}$.Using the Law of Cosines to find $PR$:$PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(30^{\circ})$$PR^2 = (10\sqrt{3})^2 + 10^2 - 2(10\sqrt{3})(10)\left(\frac{\sqrt{3}}{2}ight)$$PR^2 = 300 + 100 - 300 = 100$$PR = 10$.Since $PR = QR = 10$,the triangle is isosceles with $\angle QPR = \angle PQR = 30^{\circ}$.Then,$\angle QRP = 180^{\circ} - (30^{\circ} + 30^{\circ}) = 120^{\circ}$. Thus,statement $(B)$ is true.Area of $	riangle PQR = \frac{1}{2} 	imes PQ 	imes QR 	imes \sin(30^{\circ}) = \frac{1}{2} 	imes 10\sqrt{3} 	imes 10 	imes \frac{1}{2} = 25\sqrt{3}$.Semi-perimeter $s = \frac{10\sqrt{3} + 10 + 10}{2} = 5\sqrt{3} + 10$.Inradius $r = \frac{	ext{Area}}{s} = \frac{25\sqrt{3}}{5\sqrt{3} + 10} = \frac{5\sqrt{3}}{\sqrt{3} + 2} = \frac{5\sqrt{3}(2 - \sqrt{3})}{4 - 3} = 10\sqrt{3} - 15$. Thus,statement $(C)$ is true.Circumradius $R = \frac{abc}{4\Delta} = \frac{10\sqrt{3} 	imes 10 	imes 10}{4 	imes 25\sqrt{3}} = \frac{1000\sqrt{3}}{100\sqrt{3}} = 10$.Area of circumcircle = $\pi R^2 = \pi(10)^2 = 100\pi$. Thus,statement $(D)$ is true.Therefore,statements $(B), (C), (D)$ are true.

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