Let p, q and r be the sides opposite to the a | vedclass

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(D) Using the Sine Rule in $\Delta PQR$, we have $\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R_{c}$, where $R_{c}$ is the circumradius

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right angled

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(D) Using the Sine Rule in $\Delta PQR$, we have $\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R_{c}$, where $R_{c}$ is the circumradius of the triangle. Thus, $\sin P = \frac{p}{2R_{c}}$, $\sin Q = \frac{q}{2R_{c}}$, and $\sin R = \frac{r}{2R_{c}}$. Given the equation $r^{2} \sin P \sin Q = pq$. Substituting the values of $\sin P$ and $\sin Q$: $r^{2} \left( \frac{p}{2R_{c}} ight) \left( \frac{q}{2R_{c}} ight) = pq$ $r^{2} \frac{pq}{4R_{c}^{2}} = pq$ Since $p, q 
eq 0$, we can divide both sides by $pq$: $\frac{r^{2}}{4R_{c}^{2}} = 1$ $r^{2} = 4R_{c}^{2}$ $r = 2R_{c}$ Since $r = 2R_{c} \sin R$, we have $2R_{c} \sin R = 2R_{c}$. $\sin R = 1$ $R = 90^{\circ}$. Therefore, the triangle is a right-angled triangle.

Let $p, q$ and $r$ be the sides opposite to the angles $P, Q$ and $R$ respectively in a $\Delta PQR$. If $r^{2} \sin P \sin Q = pq$, then the triangle is

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