\(\widehat{ABC}\) nội tiếp đường tròn chắn cung \(\overparen{AC}\)
\(→\widehat{ABC}=\dfrac{1}{2}sđ\overparen{AC}\) hay \(50°=\dfrac{1}{2}sđ\overparen{AC}\)
\(↔sđ\overparen{AC}=100°\)
Ta có: \(sđ\overparen{AC}+sđ\overparen{AB}+sđ\\overparen{BC}=360°\)
hay \(100°+60°+\overparen{BC}=360°\)
\(↔160°+sđ\overparen{BC}=360°\)
\(↔sđ\overparen{BC}=200°\)
Vậy \(sđ\overparen{BC}=200°\)