Question

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

Answer 1

From the figure, we have

$cosP=\frac{adjacentside}{Hypotenuse}=\frac{3}{6}=\frac{1}{2}$

We know that, $\cos {60}^{\circ }=\frac{1}{2}$

$cosP=\cos {60}^{\circ }$

$\therefore \angle P={60}^{\circ }$

$△PQR$ is right angles at $Q$

$\therefore \angle Q={90}^{\circ }$

$In△PQR$,

$\angle P+\angle Q+\angle R={180}^{\circ }$

${60}^{\circ }+{90}^{\circ }+\angle R={180}^{\circ }$

${150}^{\circ }+\angle R={180}^{\circ }$

$\angle R={180}^{\circ }-{150}^{\circ }$

$\angle R={30}^{\circ }$

$\therefore \angle P={60}^{\circ },\angle R={30}^{\circ }$