Question

In $△ABC$, if $s\tan \frac{A}{2}=2s\tan \frac{B}{2}=3s\tan \frac{C}{2}$, then $\frac{b^2}{ac}$ is equal to

• A. $\frac{3}{5}$
• B. $\frac{13}{12}$
• C. $\frac{4}{5}$
• D. $\frac{16}{15}$

Answer 1

The correct option is D $\frac{16}{15}$

$s\tan \frac{A}{2}=2s\tan \frac{B}{2}=3s\tan \frac{C}{2}$

$⟹\frac{\tan \frac{A}{2}}{6}=\frac{\tan \frac{B}{2}}{3}=\frac{\tan \frac{C}{2}}{2}=K$

Since $\frac{A}{2}+\frac{B}{2}+\frac{C}{2}={90}^{\circ }$

$\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1$

$⟹(6\times 3+3\times 2+2\times 6)k^2=1⟹k=\frac{1}{6}$

$\therefore \sin A=\frac{2\tan \frac{A}{2}}{1+{\tan }^2\frac{A}{2}}=1$

Similarly, $\sin B=\frac{4}{5}$ and $\sin C=\frac{3}{5}$

$\therefore \frac{b^2}{ac}=\frac{{\sin }^{2}B}{\sin A\sin C}=\frac{16}{15}$