Question

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

• A. ${45}^{\circ }$
• B. ${15}^{\circ }$
• C. ${30}^{\circ }$

Answer 1

The correct option is A ${45}^{\circ }$

In $\Delta ABC$ and $\Delta PQR$,

Given,
$\angle BCA=\angle RPQ$ = ${15}^{\circ }$
$BC=PR=1km$
$AC=PQ=1.3km$
Thus, $\Delta ACB\cong \Delta QPR$ by SAS congruency.
$\Rightarrow \angle PQR=\angle BAC$ .... (CPCT)

Along the horizontal line through point A, we have
${90}^{\circ }+\angle BAC+{45}^{\circ }={180}^{\circ }$
$\angle BAC={180}^{\circ }-{45}^{\circ }-{90}^{\circ }$
$\angle BAC={45}^{\circ }$
$\Rightarrow \angle PQR=\angle BAC={45}^{\circ }$