Question

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

• A. $1/2$
• B. $2$
• C. $-2$
• D. $-1/2$

Answer 1

The correct option is D $-1/2$

$△ABD$

$\Rightarrow \sin B=\frac{AD}{BD}=\frac{AD}{\frac{a}{2}}\Rightarrow AD=\frac{a}{2}\sin B$

$\Rightarrow \cos B=\frac{AB}{BD}=\frac{AB}{\frac{a}{2}}\Rightarrow AB=\frac{a}{2}\cos B$

$△ADCapplyfinerule.$

$\Rightarrow \frac{\frac{a}{2}\sin B}{\sin (180-A-B)}=\frac{\frac{a}{2}}{\sin (A-90)}$

$\Rightarrow \sin B\cos A=-\sin (A+B)$

$\Rightarrow \sin B\cos A=-[\sin A\cos B+\cos A\sin B]$

$\Rightarrow 2\sin B\cos A+\sin A\cos B=0$

$Dividebothsidesby\cos A\cos B$

$\Rightarrow 2\tan B+\tan A=D$

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