Answer :

The diagram for the question is as -

Since QR is extended to a point S, ∠PRS becomes the exterior angle of the ΔPQR.

It is given to us that an exterior angle of a triangle is 105°.

So, ∠PRS = 105° - - - - (i)

It is known to us that -

If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

The two interior opposite angles are ∠RPQ and ∠PQR.

So, - - - - (ii)

But, in the question it is given that the interior opposite angles are equal.

⇒ ∠RPQ = ∠PQR - - - - (iii)

Substituting equation (iii) in equation (ii), we get

⇒

Substituting equation (i) in the above equation, we get

⇒ ∠RPQ =

From (iii), we have ∠RPQ = ∠PQR

⇒

Thus, each of these equal angles is . So, option (B) is correct.

Also, (Since, sum of all the angles of a triangle is 180°)

And, (By linear pair axiom)

⇒

⇒ ∠PRQ = 75°

So, we have the angles as –

∠RPQ = , ∠PQR = , ∠PRQ = 75°, and ∠PRS = 105°

Option (A) is not correct because

But, ∠PRQ is not the exterior angle. So, each of the equal angles is not .

Option (C) is not correct because . But, this is not equal to the exterior angle, ∠PRS, which is 105°. So, each of the equal angles is not equal to .

Option (D) is not correct because. The value of ∠PRQ is 75° which is not the interior angle. Thus, not equal to the exterior angle. So, each of the equal angles is not equal to 75°.

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