In the given figure PQ is a diameter of a circle with centre O and PT is a tangent at P. QT meets the circle at R. If ∠POR = 72∘, find ∠PTR.
54°
∠OQR = ∠ORQ (OR=OQ (radius))
But ∠ROP = ∠OQR + ∠ORQ (Exterior angle theorem)
72∘ = 2 ∠OQR
∠OQR = 36∘
In △QPT
∠PTQ = 180∘ - ∠PQT - ∠QPT
∠PTQ = 180∘ – 36∘ - 90∘
∠PTQ = 54∘
Therefore, ∠PTR = 54∘