Question

In the given figure, $\angle BCA=\angle RPQ={15}^{\circ }$. Then $\angle PQR$ is

• A. 45°
• B. 35°
• C. 15°
• D. 55°

Answer 1

The correct option is A

45°

In $\Delta ABC$ and $\Delta PQR,$ we have,
$\angle BCA=\angle RPQ={15}^{\circ }$ [given]
$BC=PR=1km$ [given]
$AC=PQ=1.3km$ [given]
thus, $\Delta ACB\cong \Delta QPR$ by SAS congruency.
Thus, $\angle PQR=\angle BAC$ [CPCTE]
Along the horizontal line through point A, we have,
${90}^{\circ }+\angle BAC+{45}^{\circ }={180}^{\circ }$
or, $\angle BAC={180}^{\circ }-{45}^{\circ }-{90}^{\circ }$
thus, $\angle PQR=\angle BAC={45}^{\circ }$