MCQ
- A60º
- B50º
- C80º
- D70º
✓
Answer
- 70º
Solution:
$\angle\text{BDC}=\angle\text{BAC}=60^\circ$ (Angles in the same segment of a circle)
In $\triangle\text{BDC},$ we have
$\angle\text{DBC}+\angle\text{BDC}+\angle\text{BCD}=180^\circ$ (Angle sum property of a triangle)
$\therefore50^\circ+60^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-(50^\circ+60^\circ)=(180^\circ-110^\circ)=70^\circ$
$\Rightarrow\angle\text{BCD}=70^\circ$
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