Question

In $△ABC,AB=AC,P$ and $Q$ are points on $AC$ and $AB$ respectively such that $BC=BP=PQ=AQ$. Then $\angle AQP$ is equal to (use $\pi ={180}^{\circ }$).

• A. $\frac{2\pi }{7}$
• B. $\frac{3\pi }{7}$
• C. $\frac{4\pi }{7}$
• D. $\frac{5\pi }{7}$

Answer 1

Answer: (D)

$\frac{5\pi }{7}$

Let $\angle A=\theta$
$\Rightarrow \angle APQ=\theta (\because AQ=PQ)$
$\Rightarrow \angle PQB=2\theta (\because$ Exterior angle)
$\Rightarrow \angle PQB=\angle PQB=2\theta (\because PQ=PB)$
$\Rightarrow \angle BPC=\angle A+\angle ABP=3\theta (\because$ Exterior angle)
$\Rightarrow \angle BCP=\angle BPC=3\theta (\because BP=BC)$
Also $\angle ABC=\angle ACB=3\theta (\because AB=AC)$
$\Rightarrow \angle A+\angle B+\angle C=\theta +3\theta +3\theta =7\theta =\pi \Rightarrow \theta =\frac{\pi }{7}$
$\Rightarrow \angle AQP=\pi -\angle BQP=\pi -2\theta =\pi -\frac{2\pi }{7}=\frac{5\pi }{7}$.