Question

If $CD=15,DB=9$, $AD$ bisects $\angle$ $A$, $\angle$ $ABC=90$${}^{\circ }$, then $AB$ has length

• A. $32$
• B. $18$
• C. $7$
• D. $24$

Answer 1

Answer: (B)

Let $\angle DAB=\theta$

$\Rightarrow \angle CAB=2\theta$

Let $AB=x$

$\tan 2\theta =\frac{24}{x}$ and $\tan \theta =\frac{9}{x}$

$\Rightarrow \frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{24}{x}$

$\Rightarrow \frac{\frac{18}{x}}{1-\frac{81}{x^2}}=\frac{24}{x}$

$\Rightarrow 1-\frac{81}{x^2}=\frac{3}{4}$

$\Rightarrow x^2=324$

$\Rightarrow x=18$