UPSC CDS I 2019 Math Paper

Concept:Use the property that the angle subtended by a chord at the center is twice that at the circumference.Explanation:For an equilateral triangle, each interior angle is $60^{\circ}$.The chord $BC$ subtends $\text{angle } BAC = 60^{\circ}$ at vertex $A$ on the circle.Therefore, at the center $O$, $\text{angle } BOC = 2 \times 60^{\circ} = 120^{\circ}$.Drop a perpendicular from $O$ to $BC$, meeting at $D$. This bisects $\text{angle } BOC$ and $BC$.Thus, $\text{angle } BOD = \text{angle } DOC = 60^{\circ}$.In right triangle $BOD$, $BO$ is the circumradius. For an equilateral triangle, circumradius $R = \frac{\text{side}}{\sqrt{3}}$.Given $BO = 20\sqrt{3}$ cm, we have $\sin 60^{\circ} = \frac{BD}{BO}$.So $BD = BO \cdot \sin 60^{\circ} = 20\sqrt{3} \times \frac{\sqrt{3}}{2} = 20 \times \frac{3}{2} = 30$ cm.Since $D$ is the midpoint, $BC = 2 \times BD = 60$ cm.Thus, the side of the equilateral triangle is $60$ cm.Answer:$60$ cm