If TP and TQ are the two tangents to a circle with centre O so that angle POQ = 110^{circ}, then angle PTQ is equal to:

The angle between the two tangents from an external point and the angle subtended by the line segments joining the points of contact at the centre are supplementary. Therefore, $\angle PTQ + \angle POQ = 180^{\circ}$. Substituting $\angle POQ = 110^{\circ}$, we get $\angle PTQ = 180^{\circ} - 110^{\circ} = 70^{\circ}$.