In the given figure, AB is a mirror, PQ is the incident ray and QR is the reflected ray. If ∠PQR = 108°, find ∠AQP.

Question:

In the given figure, AB is a mirror, PQ is the incident ray and QR is the reflected ray. If ∠PQR = 108°, find ∠AQP.
(a) 72°
(b) 18°
(c) 36°
(d) 54°

 

Solution:

(c) 36°

We know that angle of incidence = angle of reflection.

Then, let $\angle A Q P=\angle B Q R=x^{\circ}$

Now,

$\angle A Q P+\angle P Q R+\angle B Q R=180^{\circ} \quad[\because A Q B$ is a straight line $]$

$\Rightarrow x+108+x=180^{\circ}$

$\Rightarrow 2 x=72^{\circ}$

$\Rightarrow x=36^{\circ}$

$\therefore \angle A Q P=36^{\circ}$

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