Question

In $△ABC$, if $s\tan \frac{A}{2}=2s\tan \frac{B}{2}=3s\tan \frac{C}{2}$, then ${\left(\frac{4a}{5b}\right)}^2$ is equal to

• A. 4
• B. 1
• C. 9
• D. 25

Answer 1

Answer: (B)

1

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

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

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, $\sin B=\frac{4}{5}$

$\therefore \frac{a}{b}=\frac{\sin A}{\sin B}=\frac{5}{4}$

$⟹\frac{4a}{5b}=1⟹{\left(\frac{4a}{5b}\right)}^2=1$