Question

Let a triangle $ABC$ 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
${\left[\frac{A{A}_1\cos \frac{A}{2}+B{B}_1\cos \frac{B}{2}+C{C}_1\cos \frac{C}{2}}{\sin A+\sin B+\sin C}\right]}^2=$

• A. 4
• B. 16
• C. 25
• D. 1

Answer 1

Answer: (B)

Let the $\Delta ABC$ is equilateral.
$\Rightarrow A=B=C=60^{\circ }$
and $A{A}_1=B{B}_1=C{C}_1=$ Diameter $=4$
$\therefore {\left[\frac{A{A}_1\cos \frac{A}{2}+B{B}_1\cos \frac{B}{2}+C{C}_1\cos \frac{C}{2}}{\sin A+\sin B+\sin C}\right]}^2$
$={\left[\frac{(4)\cos 30^{\circ }+4\cos 30^{\circ }+4\cos 30^{\circ }}{\sin 60^{\circ }+\sin 60^{\circ }+\sin 60^{\circ }}\right]}^2$
$={\left[\frac{4\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)}{\frac{3\sqrt{3}}{2}}\right]}^2$
$={\left[\frac{4\times 3\frac{\sqrt{3}}{2}}{3\frac{\sqrt{3}}{2}}\right]}^2=(4{)}^2=16$