Question

$\frac{tan\angle CAB}{tan\angle CDB}$ is $\frac{1}{m}$, m is

Answer 1

In the given figure,
$\angle C={90}^{\circ }$
In $△ABC$,
$\tan \angle CAB=\frac{P}{B}=\frac{CB}{AC}$
In $△CDB$,
$\tan \angle CDB=\frac{P}{B}=\frac{CB}{CD}$
Now,
$\frac{\tan \angle CAB}{\tan \angle CDB}=\frac{\frac{CB}{AC}}{\frac{CB}{CD}}$
$\frac{\tan \angle CAB}{\tan \angle CDB}=\frac{CD}{AC}$
$\frac{\tan \angle CAB}{\tan \angle CDB}=\frac{CD}{2CD}$
$\frac{\tan \angle CAB}{\tan \angle CDB}=\frac{1}{2}$ (D is mid point of AC)