Question

If in a triangle ABC circumradius is $8$ times the inradius and $B-C=$ $\frac{2\pi }{3}$, then

• A. $\sin \frac{A}{2}$ is equal to $\frac{1}{4}$
• B. $b:c::$ $3+\sqrt{5}$ $:4$
• C. $b:c::$ $3+\sqrt{5}$ $:2$
• D. $\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{1}{32}$

Answer 1

The correct option is C $b:c::$ $3+\sqrt{5}$ $:2$

$R=8R\Rightarrow R=32R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\Rightarrow 2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{1}{16}$
or $\cos \left[\frac{B-C}{2}\right]-\cos \left[\frac{B+C}{2}\right]\sin \frac{A}{2}=\frac{1}{16}$
$\Rightarrow (\frac{1}{2}-\sin \frac{A}{2})\sin \frac{A}{2}=\frac{1}{16}\Rightarrow \sin \frac{A}{2}=\frac{1}{4}$
Again tan $\frac{B-c}{2}=\frac{b-c}{b+c}\cot \frac{A}{2}$
$\Rightarrow \sqrt{3}=\frac{b-c}{b+c}\sqrt{15}$
$\Rightarrow b:c::3+\sqrt{5}:2$