MCQ
In the given circle, $O$ is the centre and $\angle\text{BDC}=42^\circ.$ Then, $\angle\text{ACB}$ is equal to: 
- A$58^\circ$
- B$42^\circ$
- C$52^\circ$
- ✓$48^\circ$
✓
Answer
Correct option: D.
$48^\circ$
Here, $\angle\text{BDC}=\text{BAC}=42^\circ$ {Angles in same segment are equal}
now, since $AC$ is diameter so, $ABC$ forms a semi-circle, thus $\angle\text{ABC}=90^\circ$
So, in triangle ABC $\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$
$\angle\text{ACB}=180-(90+42)$
$=180^\circ-132^\circ$
$=48^\circ$
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