Question

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

• A. $1$
• B. $2$
• C. $4$
• D. $5$

Answer 1

Answer: (B)

$2$

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

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

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 $\sin C=\frac{3}{5}$

$\therefore \frac{a^2+b^2+c^2}{a^2}=\frac{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}{{\sin }^{2}A}=\frac{1+\frac{16}{25}+\frac{9}{25}}{1}=2$