Question

In $\Delta ABC,\angle B={90}^o,\angle BDC={90}^o$ and $\angle BAC={30}^o$, CX bisects $\angle C$ what is $\angle DXC$?

• A. ${30}^o$
• B. ${60}^o$
• C. ${40}^o$
• D. None

Answer 1

Answer: (B)

${60}^o$

$\angle ABC=90-30={60}^o,\angle XCD=\frac{1}{2}\times 60={30}^o$
$\therefore \angle DXC={60}^o$