In the given Figure,

As PA and PB are tangents from common external point P, we have

PA = PB

[∵ tangents drawn from an external point are equal]

∠PBA = ∠PAB

[∵ Angles opposite to equal sides are equal]

Now,

In △APB, By Angle sum Property of triangle

∠APB + ∠PBA + ∠PAB = 180°

60° + ∠PAB + ∠PAB = 180°

2 ∠PAB = 120°

∠PAB = 60°

So, We have

∠PBA = ∠PAB = ∠APB = 60°

i.e. APB is an equilateral triangle

so, we have

PA = PB = AB = 5 cm [ As PA = 5 cm ]