MCQ
In the given figure, $\triangle\text{ABC}$ and $\triangle\text{DBC}$ are inscribed in a circle such that $\angle\text{BAC}=60^\circ$ and $\angle\text{DBC}=50^\circ.$ Then $\angle\text{BCD}=?$
- A50°
- B60°
- C70°
- D80°
✓
Answer
- 70°
Solution:
Since angles in the same segment of a circle are equal.
$\angle\text{BAC}=\angle\text{BDC}=60^\circ.$
In $\triangle\text{BDC},$
$\angle\text{BDC}+\angle\text{DBC}+\angle\text{BCD}=180^\circ.$
$\Rightarrow\ 60^\circ+50^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\ \angle\text{BCD}=70^\circ$
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