Given : ∠PQR = 40° and ∠QRP = 60°  To find : ∠QOR = $$ heta$$ = ? Solution : In $$ riangle$$ PQR, => ∠ PQR + ∠ QRP + ∠QPR = $$180^circ$$ => $$40^circ+60^circ+angle QPR=180^circ$$ => $$angle QPR=180^circ-100^circ=80^circ$$ Angle subtended by an arc at the centre is double the angle subtended by it at an other point on the circle, => ∠ QOR = $$2 imes angle QPR$$ => $$ heta=2 imes 80^circ=160^circ$$ => Ans - (D)