Write 'True' or 'False' and give reasons for your answer.
If a chord $AB$ subtends an angle of $60^{\circ}$ at the centre of a circle,then the angle between the tangents at $A$ and $B$ is also $60^{\circ}$.

(B) False
Given that a chord $AB$ subtends an angle of $60^{\circ}$ at the centre $O$ of a circle.
i.e.,$\angle AOB = 60^{\circ}$.
Since $OA = OB$ (radii of the same circle),$\triangle OAB$ is an isosceles triangle.
Therefore,$\angle OAB = \angle OBA = (180^{\circ} - 60^{\circ}) / 2 = 60^{\circ}$.
Let the tangents at points $A$ and $B$ intersect at point $C$.
We know that the radius is perpendicular to the tangent at the point of contact.
So,$OA \perp AC$ and $OB \perp BC$.
Therefore,$\angle OAC = 90^{\circ}$ and $\angle OBC = 90^{\circ}$.
Now,$\angle BAC = \angle OAC - \angle OAB = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Similarly,$\angle ABC = \angle OBC - \angle OBA = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
In $\triangle ABC$,the sum of interior angles is $180^{\circ}$.
So,$\angle ACB + \angle BAC + \angle ABC = 180^{\circ}$.
$\angle ACB + 30^{\circ} + 30^{\circ} = 180^{\circ}$.
$\angle ACB = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Thus,the angle between the tangents is $120^{\circ}$,not $60^{\circ}$.