(B) False.
Since $QL$ is a tangent at $Q$,the angle between the tangent and the chord $SQ$ is equal to the angle in the alternate segment. However,using the property of the radius being perpendicular to the tangent: $\angle OQS = 90^{\circ} - \angle SQL = 90^{\circ} - 50^{\circ} = 40^{\circ}$.
In $\triangle OSQ$,$OS = OQ$ (radii of the same circle),so $\angle OSQ = \angle OQS = 40^{\circ}$.
Similarly,for the tangent $RM$ at $R$,$\angle ORS = 90^{\circ} - \angle SRM = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
In $\triangle OSR$,$OS = OR$ (radii of the same circle),so $\angle OSR = \angle ORS = 30^{\circ}$.
Therefore,$\angle QSR = \angle OSQ + \angle OSR = 40^{\circ} + 30^{\circ} = 70^{\circ}$.