PQRST is a cyclic pentagon and PT is a diameter, then $$\angle$$PQR + $$\angle$$RST is equal to

Sum of interior angles of a pentagon = ($$n$$-2)*180°

= (5-2)*180° = 540°

Since, its cyclic pentagon, => PQ = QR = RS = ST

=> $$\angle$$POQ = $$\angle$$QOR = $$\angle$$ROS = $$\angle$$SOT = $$\frac{180^{\circ}}{4}$$

= 45°

Also, OP = OQ = OR = OS = OT = radii

=> $$\angle$$OPQ = $$\frac{180^{\circ}-45^{\circ}}{2}$$ = $$\frac{135^{\circ}}{2}$$

$$\therefore$$ $$\angle$$PQR + $$\angle$$RST

= 4 * $$\frac{135^{\circ}}{2}$$ = 270°