WAEC Mathematics 2011 Paper

Concept:Apply the properties of parallel lines, the angle sum of a triangle, and supplementary angles on a straight line.Explanation:On the straight line $ON$, the angle $\angle QOP = 125^\circ$ is given.So, the adjacent angle is $\angle NOP = 180^\circ - 125^\circ = 55^\circ$.Since $MN \parallel OP$ and $ON$ is a transversal, the alternate angles are equal.Hence, $\angle MNO = \angle NOP = 55^\circ$.In $\triangle MQN$, the angle at $M$ is $\angle QMN = 65^\circ$, as shown in the diagram.Using the triangle angle sum property, $\angle QMN + \angle MNQ + \angle MQN = 180^\circ$.Substitute: $65^\circ + 55^\circ + \angle MQN = 180^\circ$.This gives $\angle MQN = 180^\circ - 120^\circ = 60^\circ$.Now, $R$, $Q$, and $N$ lie on a straight line, so $\angle MQR$ and $\angle MQN$ are supplementary.Therefore, $\angle MQR + 60^\circ = 180^\circ$.Hence, $\angle MQR = 180^\circ - 60^\circ = 120^\circ$.Answer:$\angle MQR = 120^\circ$.The correct option is B.