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?
Answer & explanation
Correct answer: option 4
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.