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 $
- B$90^\circ$
- ✓$75^\circ$
- D$135^\circ$
✓
Answer
Correct option: C.
$75^\circ$
To solve this problem we need to know that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
Here we are given that $A, B, C,$ are three points on a circle with centre $O$ such that
$\angle\text{AOB}=90^\circ$ and $\angle\text{BOC}=120^\circ.$
From
$\angle\text{AOC}=360^\circ-\angle\text{AOB}-\angle\text{BOC}$
$\Rightarrow\angle\text{AOC}=360^\circ-90^\circ-120^\circ$
$\Rightarrow\angle\text{AOC}=360^\circ-210^\circ$
$\Rightarrow\angle\text{AOC}=150^\circ$
Now, as seen earlier, the angle made by the arc $AC$ with the centre of the circle will be twice the angle it makes in any point in the remaining part of the circle.
Since the point $C$ lies on the remaining part of the circle, the angle the arc $AC$ makes with this point has to be half of the angle $'AC'$ makes with the centre. Therefore we have,
$\angle\text{ABC}=\frac{\angle\text{AOC}}{2}=\frac{150^\circ}{2}=75^\circ$
$\Rightarrow\angle\text{ABC}=75^\circ$
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