MCQ
Write the correct answer in the following: In Fig. $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ,$ then $\angle\text{CAO}$ is equal to:
- A$30^\circ .$
- B$45^\circ .$
- C$90^\circ .$
- ✓$60^\circ .$
✓
Answer
Correct option: D.
$60^\circ .$
In $\triangle\text{OAB},$ we have
$\text{OA}=\text{OB}$
[Radii of the same circle]
$\therefore\angle\text{OAB}=\angle\text{OBA}$
In triangle $OAB,$ we have
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$
$\therefore2\angle\text{OAB}=(180^\circ-\angle\text{AOB})$
$=(180^\circ-90^\circ)=90^\circ$ [$\because$ sum of angles of $\triangle$ is $180^\circ$]
$\Rightarrow\angle\text{OAB}=\frac{1}{2}\times90^\circ=45^\circ$
Also, $\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}=\frac{1}{2}\times90^\circ=45^\circ$
Now, in $\triangle\text{CAB},$ we have
$\angle\text{CAB}=180^\circ-(\angle\text{ABC}+\angle\text{ACB})$
$=180^\circ-(30^\circ+45^\circ)=105^\circ$
Now, $\angle\text{CAO}=\angle\text{CAB}-\angle\text{OAB}$
$\Rightarrow\angle\text{CAO}=105^\circ-45^\circ=60^\circ$
Hence, $(d)$ is the correct answer.
$\text{OA}=\text{OB}$
[Radii of the same circle]
$\therefore\angle\text{OAB}=\angle\text{OBA}$
In triangle $OAB,$ we have
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$
$\therefore2\angle\text{OAB}=(180^\circ-\angle\text{AOB})$
$=(180^\circ-90^\circ)=90^\circ$ [$\because$ sum of angles of $\triangle$ is $180^\circ$]
$\Rightarrow\angle\text{OAB}=\frac{1}{2}\times90^\circ=45^\circ$
Also, $\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}=\frac{1}{2}\times90^\circ=45^\circ$
Now, in $\triangle\text{CAB},$ we have
$\angle\text{CAB}=180^\circ-(\angle\text{ABC}+\angle\text{ACB})$
$=180^\circ-(30^\circ+45^\circ)=105^\circ$
Now, $\angle\text{CAO}=\angle\text{CAB}-\angle\text{OAB}$
$\Rightarrow\angle\text{CAO}=105^\circ-45^\circ=60^\circ$
Hence, $(d)$ is the correct answer.
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