Ans: B
In $\Delta OBC$,
In $\Delta OBC$,
$\begin{array}{ll}
OB=OC & \text{(radii)} \\
\angle OCB = \angle OBC & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle OCB = \angle 50^\circ & \\
\therefore \angle ACB = 30^\circ
\end{array}$
Hence, we have
$\begin{array}{ll}
\angle BOA = 2 \times \angle ACB & \text{($\angle$ at centre twice $\angle$ at }\unicode{x2299}^{ce}) \\
\angle BOA = 2 \times 30^\circ \\
\angle BOA = 60^\circ
\end{array}$
