Question

In the given figure; $\angle C={90}^0$ and $D$ is midpoint of $AC$. If $\frac{\tan \angle CAB}{\tan \angle CDB}$ is $\frac{1}{m}$, then $m$ is

• A. 2
• B. 3
• C. 1
• D. 7

Answer 1

The correct option is A 2

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)