Question

From the given figure, the value of $\frac{\tan \angle ABC}{\tan \angle DBC}$ is:

• A. 1
• B. 0
• C. 2
• D. 4

Answer 1

Answer: (C)

2

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,

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

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

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

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