MCQ
If $A , B, C$ are three points on a circle with centre $O$ such that $\angle\text{AOB} = 90^\circ$ and $\angle\text{BOC} = 120^\circ,$ then $\angle\text{ABC} =$
- A$60^\circ $
- ✓$75^\circ$
- C$90^\circ$
- D$135^\circ$
✓
Answer
Correct option: B.
$75^\circ$
$\angle\text{AOC}=\angle\text{AOB}+\angle\text{BOC}$
$=90^\circ+120^\circ=210^\circ$
$\angle\text{COA}=360^\circ-210^\circ=150^\circ$
If arc $\widehat{\text{COA}}$ makes $150^\circ$ at centre, then it will make half angle of the centre at circumference.
$\Rightarrow\angle\text{CBA}$ or $\angle\text{ABC}=\frac{150^\circ}{2}=75^\circ$
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