In a non-right-angled triangle triangle PQR, | vedclass

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(C) Using the sine rule, $\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R_{c} = 2(1) = 2$.Given $p=\sqrt{3}, q=1$, we have $\sin P = \frac

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$2, 3, 4$

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(C) Using the sine rule, $\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R_{c} = 2(1) = 2$.Given $p=\sqrt{3}, q=1$, we have $\sin P = \frac{\sqrt{3}}{2}$ and $\sin Q = \frac{1}{2}$.Since $p > q$, $P > Q$. Possible values for $P$ are $60^{\circ}$ or $120^{\circ}$, and for $Q$ are $30^{\circ}$ or $150^{\circ}$.If $P=60^{\circ}, Q=30^{\circ}$, then $R=90^{\circ}$ (rejected as it is non-right-angled).If $P=120^{\circ}, Q=30^{\circ}$, then $R=30^{\circ}$. Thus, $	riangle PQR$ is isosceles with $q=r=1$.Area $\Delta = \frac{1}{2}qr \sin P = \frac{1}{2}(1)(1)\sin 120^{\circ} = \frac{\sqrt{3}}{4}$.Semi-perimeter $s = \frac{\sqrt{3}+1+1}{2} = \frac{\sqrt{3}+2}{2}$.Inradius $r_{in} = \frac{\Delta}{s} = \frac{\sqrt{3}/4}{(\sqrt{3}+2)/2} = \frac{\sqrt{3}}{2(2+\sqrt{3})} = \frac{\sqrt{3}}{2}(2-\sqrt{3})$. (Option $2$ is correct).Length of median $RS = \frac{1}{2}\sqrt{2p^2+2q^2-r^2} = \frac{1}{2}\sqrt{2(3)+2(1)-1} = \frac{\sqrt{7}}{2}$. (Option $3$ is correct).$PE$ is the altitude to $QR$. $PE = q \sin R = 1 \sin 30^{\circ} = \frac{1}{2}$.$O$ is the centroid of $	riangle PQR$ (since $S$ is midpoint of $PQ$ and $PE$ is altitude, $O$ is intersection of median $RS$ and altitude $PE$ is not generally the centroid, but here $PQR$ is isosceles with $q=r$, so altitude $PE$ is also the median). Thus $O$ is the centroid, $OE = \frac{1}{3}PE = \frac{1}{6}$. (Option $4$ is correct).Area $	riangle SOE = \frac{1}{6} \Delta = \frac{1}{6} \cdot \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{24}$. (Option $1$ is incorrect).

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