In the given figure, AB is a mirror, PQ is the incident ray and QR, the reflected ray. If `anglePQR=112^(@)`, find `anglePQA`.

Options

A. `35^(@)`
B. `34^(@)`
C. `36^(@)`
D. `30^(@)`

Answer

B

Explanation

When a ray falls on a mirror, it is reflected and angle of incidence = angle of reflection `=x^(@)` (say).
QM is drawn normal to AB and therefore,
angle of incidence `=anglePQM`,
angle of relection `=angleMQR`
and `angleAQM=90^(@)`.
Now, `anglePQM+angleMQR=anglePQR=112^(@)` (given)
`:. 2anglePQM=112^(@)`
`:. anglePQM=56^(@)`.
So, `anglePQA=angleAQM-anglePQM=90^(@)-56^(@)=34^(@)`.