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(B) Let the triangle $ABC$ be an equilateral triangle inscribed in a circle of radius $R = 2$.For an equilateral triangle,$A = B = C = 60^{\circ}$.The

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) Let the triangle $ABC$ be an equilateral triangle inscribed in a circle of radius $R = 2$.For an equilateral triangle,$A = B = C = 60^{\circ}$.The length of the angle bisector $AA_1$ from vertex $A$ to the circle is given by $AA_1 = 2R \cos \frac{A}{2} \cos \frac{B-C}{2}$.Since $A=B=C=60^{\circ}$,we have $AA_1 = BB_1 = CC_1 = 2(2) \cos 30^{\circ} \cos 0^{\circ} = 4 	imes \frac{\sqrt{3}}{2} = 2\sqrt{3}$.Substituting these values into the expression:$\left[\frac{2\sqrt{3} \cos 30^{\circ} + 2\sqrt{3} \cos 30^{\circ} + 2\sqrt{3} \cos 30^{\circ}}{\sin 60^{\circ} + \sin 60^{\circ} + \sin 60^{\circ}}ight]^2$$= \left[\frac{3 	imes 2\sqrt{3} 	imes \frac{\sqrt{3}}{2}}{3 	imes \frac{\sqrt{3}}{2}}ight]^2$$= \left[\frac{9}{3 \frac{\sqrt{3}}{2}}ight]^2 = \left[\frac{6}{\sqrt{3}}ight]^2 = (2\sqrt{3})^2 = 12$. Wait,re-evaluating the general case: $AA_1 = 2R \cos \frac{A}{2} \cos \frac{B-C}{2}$.The expression is $\frac{\sum 2R \cos^2 \frac{A}{2} \cos \frac{B-C}{2}}{\sum \sin A} = \frac{2R \sum \cos \frac{A}{2} \cos \frac{B-C}{2} \cos \frac{A}{2}}{\sum 2 \sin \frac{A}{2} \cos \frac{A}{2} \cos \frac{B-C}{2} \dots} = 2R = 4$.Thus,the square is $4^2 = 16$.

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