Question

$△ABC$ is a right angled triangle, where $\angle B={90}^{\circ }$ and $BD\perp AC$. If $BD=8cm,AD=4cm$; then $CD$ is

• A. $24cm$
• B. $16cm$
• C. $32cm$
• D. $8cm$

Answer 1

Answer: (B)

$16cm$

Let $\angle C=\theta$

In $△CBD,\tan \theta =\frac{BD}{DC}=\frac{8}{DC}$

$△BDA,\tan \angle ABD=\frac{AD}{BD}=\frac{4}{8}=\frac{1}{2}$

Equating the two equations

$\frac{8}{DC}=\frac{1}{2}$

$DC=16cm$