Question

In $△$ PQR, right angled at Q, PQ = 12 cm and PR = 12$\sqrt{2}$ cm. Determine $\angle$QPR and $\angle$ PRQ.

• A. ${60}^{\circ }$ , ${30}^{\circ }$
• B. ${45}^{\circ }$ , ${45}^{\circ }$
• C. ${30}^{\circ }$ , ${60}^{\circ }$
• D. ${22.5}^{\circ }$ , ${67.5}^{\circ }$

Answer 1

The correct option is B

${45}^{\circ }$ , ${45}^{\circ }$

We know, PQ = 12cm

In $△$ PQR, Sin R = $\frac{PQ}{PR}$ = $\frac{12}{\sqrt{2}}$ = $\frac{1}{\sqrt{2}}$

When R = ${45}^{\circ }$, then only sin R = $\frac{1}{\sqrt{2}}$

Apply Pythagoras theorem in $△$PQR

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

$(12\sqrt{2}{)}^2$= ${12}^2$+ $Q{R}^2$

288 = 144 + $Q{R}^2$

$Q{R}^2$ = 144

QR = 12 cm

In $△$ PQR, tan P = $\frac{QR}{PQ}$ = 1

When P = ${45}^{\circ }$, then only tan P = 1

$\Rightarrow$ $\angle$P = $\angle$R = ${45}^{\circ }$