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