Question

$AB\perp BC,BD\perp AC$ and $CE$ bisects $\angle C,\angle A={30}^{\circ }$. Then what is $\angle CED$?

• A. ${30}^o$
• B. ${60}^o$
• C. ${45}^o$
• D. ${65}^o$

Answer 1

Answer: (B)

${60}^o$

In $\Delta ABC$,

$\angle ABC+\angle BAC+\angle ACB={180}^{\circ }$
$\Rightarrow {90}^{\circ }+{30}^{\circ }+\angle ACB={180}^{\circ }$

$\Rightarrow \angle ACB={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

Now $CE$ bisects $\angle C\Rightarrow \angle ECD={30}^{\circ }$

Also, $BD\perp AC\Rightarrow \angle EDC={90}^{\circ }$
$\therefore$ In $\Delta DEC,$

$\angle ECD+\angle CED={180}^{\circ }$

$\Rightarrow {30}^{\circ }+{90}^{\circ }+\angle CED={180}^{\circ }$

$\Rightarrow \angle CED={180}^{\circ }={180}^{\circ }-{120}^{\circ }={60}^{\circ }$.