In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T.

AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∴∠APB = 90∘
By using alternate segment theorem
We have ∠APB = ∠PAT = 30∘ 
Now, in △APB
∠BAP + ∠APB + ∠BAP = 180∘      (Angle sum property of triangle)
⇒ ∠BAP = 180∘ − 90∘ − 30∘ = 60∘
Now, ∠BAP = ∠APT + ∠PTA        (Exterior angle property)
⇒ 60∘ = 30∘ + ∠PTA
⇒ ∠PTA = 60∘ − 30∘ = 30∘
We know that sides opposite to equal angles are equal.
∴ AP = AT
In right triangle ABP

$\sin \angle \mathrm{ABP}=\frac{\mathrm{AP}}{\mathrm{BA}}$

$\Rightarrow \sin 30^{\circ}=\frac{\mathrm{AT}}{\mathrm{BA}}$

$\Rightarrow \frac{1}{2}=\frac{\mathrm{AT}}{\mathrm{BA}}$

$\therefore \mathrm{BA}: \mathrm{AT}=2: 1$