In, △OQR,
∵OQ=OR (Radius)
(Angles opposite to equal sides are equal.)
∴∠OQR=∠ORQ
By exterior angle property,
∠POR=∠OQR+∠ORQ⇒78∘=2∠OQR⇒∠OQR=39∘∠PQR=∠OQR=39∘
We know that, tangent at any point is perpendicular to the radius at the point of contact.
∴∠TPQ=90∘
Now, consider △TPQ,
∠TPQ+∠PQR+∠QTP=180∘
∠QTP=180∘−39∘−90∘
∴∠QTP=∠PTR=51∘