In circle with centre O and radius 13 cm, a chord AB is drawn. Tangents at A and B intersect at P such that ∠APB = 60°. If Distance of AB from the centre O is 5 cm, then what is the length (in cm) of AP?

24

Since OM = 5 (given)

OA = 13 (given)

Using pythagoras theorem

AM = 12

As AB = 2 AM (perpendicular from the center bisect the chord in equal part)

AB = 2 x 12 = 24

In \(\Delta \)PAB,

\(\angle\)APB = \({60}^\circ\)

As PA = PM (as two tangents are equal from the same point)

So, \(\Delta \)PAB will be an equilateral triangle.

So, AP = 24 cm

Therefore, AP is 24 cm.