Question

If the median of a triangle $ABC$ through $A$ is perpendicular to $AB$ then $\frac{\tan A}{\tan B}$ has the value equal to

• A. $\frac{1}{2}$
• B. $2$
• C. $-2$
• D. $\frac{-1}{2}$

Answer 1

The correct option is C $-2$

According to sine rule:-

$\frac{\sin C}{c}=\frac{\sin A}{a}=\frac{\sin B}{b}$

Now j$\frac{a}{2}\cos B=C$ and $C=\pi -(A+B)$

$\therefore \frac{\sin [\pi -(A+B)]}{\frac{a}{2}\cos B}=\frac{\sin A}{a}$

$\therefore \frac{2\sin (A+B)}{\cos B}=\sin A$

$\frac{2(\sin A\cos B+\sin B\cos A)}{\sin A\cos B}=1$

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

$\therefore 2\frac{\tan B}{\tan A}=-1$

$\therefore \frac{\tan A}{\tan B}=-2$