Question

Let $ABC$ be an acute-angled triangle and let $D$ be the midpoint of $BC$. If $AB=AD$, then $\frac{\tan B}{\tan C}$ equals to

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

Answer 1

The correct option is D $3$

In $△ADB,$

$\angle ABD=\angle ADB$ ........ [Angles opposite to equal sides are equal]

$\therefore \angle ADC=\pi -\angle ABD$

Also, $\angle

Using $2^{\text{nd}}$ formula in $m-n\cot$ theorem

Here m=1 and n=1

$(1+1)\cot (\pi -B)$ = $1\cot B-1\cot C$

$3\cot B=\cot C$

$\frac{\tan B}{\tan C}=3$