Question
On a semi-circle with AB as diameter, a point C is taken, so that $\text{m}\big(\angle\text{CAB}\big)=30^\circ.$ Find $\text{m}\big(\angle\text{ACB}\big)$ and $\text{m}\big(\angle\text{ABC}\big).$
✓
Answer
We have, $\angle\text{CAB}=30^\circ$$\therefore\angle\text{ACB}=90^\circ$ [Angle in semicircle]
In $\triangle\text{ABC},$ by angle sum property
$\angle\text{CAB}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$
$\Rightarrow30^\circ+90^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-90^\circ-30^\circ=60^\circ$
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