Question

In the given figure, $\angle C={90}^{\circ }$ and D is the mid point of AC. Then $\frac{cot\angle CAB}{cot\angle CDB}=$ ___

• A. $\frac{1}{2}$
• B. 0
• C. 2
• D. 1

Answer 1

The correct option is C

2

Let AD = DC = x

and BC = y

$cot\angle CAB=\frac{AC}{BC}$

$=\frac{2X}{Y}$

$cot\angle CDB=\frac{DC}{BC}$

$=\frac{X}{Y}$

$\therefore \frac{cot\angle CAB}{cot\angle CDB}=\frac{\frac{2x}{y}}{\frac{x}{y}}=2$