MCQ
In the given figure, equilateral $\triangle\text{ABC}$ is inscribed in a circle and $ABCD$ is a quadrilateral, as shown. Then, $\angle\text{BDC}=?$
- A$150^\circ$
- ✓$120^\circ$
- C$60^\circ$
- D$90^\circ$
✓
Answer
Correct option: B.
$120^\circ$
$\triangle\text{ABC}$ is an equilateral triangle so $\angle\text{BAC}=60^\circ$
In cyclic quadrilateral $ABCD,$ we have:
$\angle\text{BDC}+\angle\text{BAC}=180^\circ$
$\Rightarrow\angle\text{BDC}+60^\circ=180^\circ$
$\therefore\angle\text{BDC}=(180^\circ-60^\circ)=120^\circ$
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