Question

Let $ABC$ be an acute-angled triangle and let $D$ be the mid-point of $BC$. If $AB=AD$, then $\tan (B)/\tan (C)$ equals

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

Given, in $\Delta ABC,D$ is mid-point of $BC$ and $AB=AD$
$\therefore \angle B=\angle ADB$
$\theta =\pi -\angle ADB=\pi -B$
$BD=DC=1:1$

Apply $(m-n)$ theorem,
$(m+n)\cot \theta =n\cot B-m\cot C$
$\Rightarrow (1+1)\cot (\pi -B)=\cot B-\cot C$
$[\because m=n=1]$
$\Rightarrow -2\cot B=\cot B-\cot C$
$[\because \cot (\pi -B)=-\cot B]$
$\Rightarrow 3\cot B=\cot C$
$\Rightarrow \frac{\tan B}{\tan C}=3$