Question 42

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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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