MCQ
In the given figure, $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ.$ Then, $\angle\text{CAO}=?$ 
- A$30^\circ$
- B$45^\circ$
- ✓$60^\circ$
- D$90^\circ$
✓
Answer
Correct option: C.
$60^\circ$
$\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(90^\circ)$
$\Rightarrow\ \angle\text{ACB}=45^\circ$
$\angle\text{COA}=2\angle\text{CBA}=2(30^\circ)=60^\circ$
Since $AOD$ is a straight line,
$\therefore\ \angle\text{COD}+\angle\text{AOC}=180^\circ$
$\therefore\ \angle\text{COD}+60^\circ=180^\circ$
$\therefore\ \angle\text{COD}=120^\circ$
$\Rightarrow\ \angle\text{CAO}=\frac{1}{2}\angle\text{COD}=\frac{1}{2}\times120^\circ=60^\circ$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(90^\circ)$
$\Rightarrow\ \angle\text{ACB}=45^\circ$
$\angle\text{COA}=2\angle\text{CBA}=2(30^\circ)=60^\circ$
Since $AOD$ is a straight line,
$\therefore\ \angle\text{COD}+\angle\text{AOC}=180^\circ$
$\therefore\ \angle\text{COD}+60^\circ=180^\circ$
$\therefore\ \angle\text{COD}=120^\circ$
$\Rightarrow\ \angle\text{CAO}=\frac{1}{2}\angle\text{COD}=\frac{1}{2}\times120^\circ=60^\circ$
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