Question 25

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Jambmaths question: 

An equilateral triangle of sides $\sqrt{3}$is inscribed in a circle. Find the radius of the circle.

Option A: 

$\tfrac{2}{3}$cm

Option B: 

1cm

Option C: 

2cm

Option D: 

3cm

Jamb Maths Solution: 

$\begin{align}  & AD=\sqrt{3}\sin {{60}^{o}}=\sqrt{3}\times \frac{\sqrt{3}}{2}=\frac{3}{2} \\ & \angle OBD={{30}^{o}}\text{        }\!\!\{\!\!\text{ Angular bisector of }\angle ABC\} \\ & \frac{OD}{BD}=\tan {{30}^{o}} \\ & OD=BD\tan {{30}^{o}} \\ & OD=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{3}}=\frac{1}{2} \\ & OA=AD-DO=\frac{3}{2}-\frac{1}{2}=\frac{2}{2}=1 \\ & AO=1cm \\\end{align}$

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