It is given that AB is …

We have, AB = 16 cm
{tex} \therefore{/tex} AL = BL = 8 cm [Perpendicular from the centre of the circle to a chord bisects it]
In {tex}\triangle{/tex}OLB, we have,
OB² = OL² + LB²
{tex} \Rightarrow{/tex} 10² = OL² + 8²
{tex} \Rightarrow{/tex} OL² = 100 - 64 = 36
{tex} \Rightarrow{/tex} OL = 6 cm
Let PL = x and PB = y. Then, OP = (x + 6) cm.
Now, {tex} \angle PBO = 90^\circ{/tex}  as tangent makes a right angle with the radius of the circle at the point of contact.
In {tex}\triangle{/tex}'s PLB and OBP, we have,
PB² = PL² + BL² and OP² = OB² + PB²
{tex} \Rightarrow{/tex} y² = x² + 64 and (x + 6)² = 100 + y² [Substituting the value of y² in second equation]
{tex} \Rightarrow{/tex} (x + 6)² = 100 + x² + 64
{tex} \Rightarrow{/tex} 12x = 128
{tex}\Rightarrow x = \frac{{32}}{3}cm{/tex}
{tex} \therefore{/tex} y² = x² + 64
{tex} \Rightarrow {y^2} = {\left( {\frac{{32}}{3}} \right)^2} + 64 = \frac{{1600}}{9}{/tex}
{tex} \Rightarrow y = \frac{{40}}{3}cm{/tex}
Therefore, {tex} PA = PB = \frac{{40}}{3}cm{/tex}