Question

In $△PQR$, right-angled at $Q$, $PQ=3$ $cm$ and $PR=6$ $cm$. Detrmine $\angle QPR$ AND $\angle PRQ$.

Answer 1

$P{R}^2=P{Q}^2+Q{R}^2$

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

$QR=3\sqrt{3}$

$cos\left(\angle QPR\right)=\frac{1}{2}$

$\angle QPR={60}^0$

$cos\left(\angle QRP\right)=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}$

$\angle \left(QRP\right)={30}^0$