Question

In $\Delta PQR$, right -angled at $Q,PQ=3cm$ and $PR=6cm$. Determine $\angle QPR$ and $\angle PQR$.

Answer 1

In $\Delta$PQR; LQ$={90}^o$

PQ$=3$cm; PR$=6$cm

QR$=\sqrt{6^2-3^2}=3\sqrt{3}$

$\sin \left(LQPR\right)=\frac{QR}{PR}$

$=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}$

So, LQPR$={60}^o\Rightarrow$ LQRP$={30}^o$.