Question

In the figure, AB and BC are two plane mirrors
perpendicular to each other. Prove that the incident
ray PQ is parallel to ray RS.
​

Answer 1

Given:

In the figure, AB and BC are two plane mirrors perpendicular to each other.

To prove:

Ray PQ || Ray RS

Proof:

Let angle of incidence of ray PQ be $\theta$.

$\therefore \angle PQO =\theta$

Now , angle of incidence is equal to angle of reflection.

$\therefore \angle OQR =\theta$

$\therefore \angle PQR =2\theta$

Now, since OQ $\perp$ BC , we can say:

$\angle RQB = {90}^{\circ}-\theta$

Now , in ∆RQB ,

$\angle RQB +\angle QRB + \angle QBR ={180}^{\circ}$

$=>({90}^{\circ}-\theta)+\angle QRB + \angle QBR ={180}^{\circ}$

$=>\angle QRB=\theta$

Again OR $\perp$ AB , we can say:

$\angle ORQ = {90}^{\circ}-\theta$

Again due to Law of reflection:

$\angle QRS = 2 \times ({90}^{\circ}-\theta)$

$=>\angle QRS = {180}^{\circ}-2\theta$

So, $\angle QRS + \angle PQS = {180}^{\circ}$

Hence , sum of internal angles between the transversal is 180°

So, PQ || RS and RQ is transversal.

[Hence proved]

Answer 2

Step-by-step explanation:

Given: Two plane mirrors m and n, perpendicular to each other. CA is incident ray and BD is reflected ray.

To Prove: CA∥DB

Construction: OA and OB are perpendiculars to m and n respectively.

Proof:

∵m⊥n,OA⊥m and OB⊥n

∴∠AOB=90  

(Lines perpendicular to two perpendicular lines are also perpendicular.)

In ΔAOB

∠AOB+∠OAB+∠OBA=180  

⇒90  +∠2+∠3=180  

⇒∠2+∠3=90  

⇒2(∠2+∠3)=180  

 (Multiplying both sides by 2)

⇒2(∠2)+2(∠3)=180  

⇒∠CAB+∠ABD=180  

(Angle of incidence = Angle of reflection)

∴∠1=∠2 and ∠3=∠4)

⇒CA∥BD    (∠CAB & ∠ABD form a pair of consecutive interior angles and are supplementary)

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