Two tangents PA and PBare drawn to a circle with centre O from an external point P. If $$\angle OAB = 30^\circ$$, then $$\angle APB$$ is:

PA & PB are the tangents to a circle, with Centre O from a point P outside it.
We know that the tangents to a circle from an external point are equal in length so PA= PB.
PA =PB

∠PBA = ∠PAB
[Angles opposite to the equal sides of a triangle are equal.]

∠APB+ ∠PBA +∠PAB= 180°
[Sum of the angles of a triangle is 180°]

x° + ∠PAB +∠PAB = 180°
[∠PBA = ∠PAB]
x° + 2∠PAB = 180°
∠PAB =½(180° - x°)
∠PAB =90° - x°/2
∠OAB +∠PAB=90°
∠OAB =90° - ∠PAB
∠OAB =90° - (90° - x°/2)
∠OAB =90° - 90° + x°/2
∠OAB = x°/2
∠OAB = ∠APB /2
∠OAB = 1/2∠APB
∠APB = 2∠OAB
∠APB =$$2 \times 30^{0}$$