In the given figure, O is the centre of the circle, ∠PQR = 100° and ∠STR = 105°. What is the value (in degrees) of ∠OSP?

PQRS is a cyclic quadrilateral.

∠PQR = 100°

∴ ∠PSR = 180° – 100° = 80°

∠STR = 105°

∴ ∠SOR = 360° – 2 × 105°

= 360° – 210° = 150º

∴ ∠OSR = ∠ORS

$=\frac{1}{2}\left(180^{\circ}-150^{\circ}\right)=15^{\circ}$

$\therefore \angle \mathrm{PSO}=80^{\circ}-15^{\circ}=65^{\circ}$