In the given figure, there are two concentric circles with center O such that AP is tangent to the bigger circle and  AB is tangent to the smaller circle. If ∠APB=∠ABP=30∘,OA=3 cm and OP =5 cm, then, radius of the smaller circle is

The problem seems to be missing the figure, but based on the description, we can solve it using the properties of right triangles and circles.

1. Since AP is tangent to the bigger circle and AB is tangent to the smaller circle, we know that ∠OAP and ∠OBA are right angles (90 degrees). This is because a radius perpendicular to a tangent at the point of contact forms a right angle.

2. Now, we have two right triangles: ∆OAP and ∆OBA. In both of these triangles, we know one angle (30 degrees) and the length of the hypotenuse (OA = 3 cm for ∆OBA and OP = 5 cm for ∆OAP).

3. We can use the sine of the known angle to find the opposite side in each triangle. In ∆OAP, the opposite side is AP, and in ∆OBA, the opposite side is AB.

4. The sine of 30 degrees is 0.5. So, AP = 0.5 * OP = 0.5 * 5 cm = 2.5 cm, and AB = 0.5 * OA = 0.5 * 3 cm = 1.5 cm.

5. The radius of the smaller circle is OB, which is equal to OA - AB = 3 cm - 1.5 cm = 1.5 cm.

So, the radius of the smaller circle is 1.5 cm.