Question

In the given figure; $\angle C={90}^{\circ }$ and D is midpoint of AC. Find $\frac{tan\angle ABC}{tan\angle DBC}$

Answer 1

In the given figure,
$\angle C={90}^{\circ }$
In $△ABC$,
$\tan \angle ABC=\frac{P}{B}=\frac{AC}{BC}$
In $△CDB$,
$\tan \angle DBC=\frac{P}{B}=\frac{CD}{BC}$
Now,
$\frac{\tan \angle ABC}{\tan \angle DBC}=\frac{\frac{AC}{BC}}{\frac{CD}{BC}}$

$\frac{\tan \angle ABC}{\tan \angle DBC}=\frac{AC}{CD}$

$\frac{\tan \angle ABC}{\tan \angle DBC}=\frac{2CD}{CD}$

$\frac{\tan \angle ABC}{\tan \angle DBC}=2$ (D is mid point of AC)