Question

Let a triangle $A B C$ be inscribed in a circle of radius $2$ units. If the $3$ bisectors of the angles $A, B$ and $C$ are extended to cut the circle at $A_{1}, B_{1}$ and $C$, respectively, then the value of
$\left[\frac{AA_{1}\cos \frac{A}{2} +BB_{1} \cos \frac{B}{2}+CC_{1} \cos \frac{C}{2}}{\sin A +\sin B +\sin C}\right]^{2} = $

• A. 4
• B. 16
• C. 25
• D. 1

Answer 1

Answer: (B)

Let the $\Delta A B C$ is equilateral.
$\Rightarrow A=B=C=60^{\circ}$
and $A A_{1}=B B_{1}=C C_{1}=$ Diameter $=4$
$\therefore \left[\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin B+\sin C}\right]^{2}$
$=\left[\frac{(4) \cos 30^{\circ}+4 \cos 30^{\circ}+4 \cos 30^{\circ}}{\sin 60^{\circ}+\sin 60^{\circ}+\sin 60^{\circ}}\right]^{2}$
$=\left[\frac{4\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)}{\frac{3 \sqrt{3}}{2}}\right]^{2}$
$=\left[\frac{4 \times 3 \frac{\sqrt{3}}{2}}{3 \frac{\sqrt{3}}{2}}\right]^{2}=(4)^{2}=16$