Correct option is (B) 30°
Given that \(\angle P O Q=120^{\circ}\)
\(\angle O P T=\angle O QT=90^{\circ}\)
[\(\because\)A tangent at any point of a Circle is perpendicular to the radius at the point of contact]
\(\angle P O T =\frac{1}{2} \angle P O Q \)
\(=\frac{1}{2} \times 120^{\circ} \)
\(=60^{\circ}\)
In \(\triangle P O T\),
\(\angle O P T+\angle P O T+\angle O T P =180^{\circ}\)
\(90^{\circ}+60^{\circ}+\angle O T P =180^{\circ}\)
\(\angle O T P =180^{\circ}-150^{\circ}\)
\(=30^{\circ}\)
