MCQ
In the given figure, $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ.$ Then, $\angle\text{CAO}=?$ 
- A$45^\circ$
- B$90^\circ$
- C$30^\circ$
- ✓$60^\circ$
We have:
$\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}=(\frac{1}{2}\times90^\circ)=45^\circ$
$\Rightarrow\angle\text{ACB}=45^\circ$
$\angle\text{COA}=2\angle\text{CBA}=(2\times30^\circ)=60^\circ$
$\therefore\angle\text{COD}=180^\circ-\angle\text{COA}=(180^\circ-60^\circ)=120^\circ$
$\Rightarrow\angle\text{CAO}=\frac{1}{2}\angle\text{COD}=(\frac{1}{2}\times120^\circ)=60^\circ$
$\Rightarrow\angle\text{CAO}=60^\circ$
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