waecmaths question:
The diagram is a circle of radius $\left| OQ \right|=4cm$ . $\overline{TR}$ is a tangent to the circle at R. If $\angle TPO={{120}^{{}^\circ }}$ find $\left| PQ \right|$
Option A:
2.32cm
Option B:
1.84cm
Option C:
0.62cm
Option D:
0.26cm
waecmaths solution:
Draw a line from O to ROQ = OR {Radius of circle}\[\begin{align} & \angle TPO={{120}^{{}^\circ }} \\ & \angle OPR={{180}^{\circ }}-{{120}^{{}^\circ }}={{60}^{\circ }}\text{ }\!\!\{\!\!\text{ }\angle \text{s on straight line }\!\!\}\!\!\text{ } \\ & \sin {{60}^{\circ }}=\frac{\left| RO \right|}{\left| PO \right|} \\ & \frac{\sqrt{3}}{2}=\frac{4}{\left| PO \right|} \\ & \left| PO \right|=\frac{4\times 2}{\sqrt{3}}=\frac{8}{\sqrt{3}}=4.62 \\ & \left| PQ \right|=\left| PO \right|-\left| QO \right|=(4.62-4)=0.62 \\\end{align}\]
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