WAEC Mathematics 2012 Paper

Concept:A tangent is perpendicular to the radius at the point of contact. Use trigonometry in the right triangle formed by the centre, the point of tangency, and the external point.Explanation:Let $O$ be the centre of the circle.Since line $TR$ is tangent to the circle at $R$, we have $OR \perp TR$.As $P$ lies on the tangent line $TR$, it follows that $OR \perp PR$.Therefore, $\triangle OPR$ is right-angled at $R$.Given $\angle TPO = 120^\circ$, and $PT$ is the opposite ray of $PR$, the acute angle at $P$ in $\triangle OPR$ is$$\angle OPR = 180^\circ - 120^\circ = 60^\circ.$$The radius of the circle is $OR = 4$ cm.In $\triangle OPR$,$$\sin 60^\circ = \frac{OR}{OP} = \frac{4}{OP}.$$So,$$OP = \frac{4}{\sin 60^\circ} = \frac{4}{\frac{\sqrt{3}}{2}} = \frac{8}{\sqrt{3}} \approx 4.62 \text{ cm}.$$Since $Q$ is the point where $OP$ meets the circle, $OQ = OR = 4$ cm.Hence,$$PQ = OP - OQ \approx 4.62 - 4 = 0.62 \text{ cm}.$$Answer:$PQ = 0.62$ cm (Option C).