In the given figure, PA is the tangent to the circle with centre O nsuch that OA = 10 cm, AB = 8 cm and AB OP . Find…

In right angled $\triangle OBA$, $OB = \sqrt{(10)^2 - (8)^2} = 6$ cm (1/2 Mark)
Let $\angle AOB = \theta$
So, $\tan \theta = \frac{8}{6}$ --- (i) (1/2 Mark)
In right angled $\triangle OAP$
$\frac{AP}{10} = \tan \theta$
$\therefore \frac{AP}{10} = \frac{8}{6}$ [using (i)] (1/2 Mark)
$\Rightarrow AP = \frac{40}{3}$ cm (1/2 Mark)
$\therefore OP = \sqrt{(\frac{40}{3})^2 + (10)^2} = \frac{50}{3}$ cm (1/2 Mark)
$PB = OP - OB = \frac{50}{3} - 6 = \frac{32}{3}$ cm or $10.6$ cm (1/2 Mark)
Alternate solution:
In right angled $\triangle OBA$,
$OB = \sqrt{(10)^2 - (8)^2} = 6$ cm (1/2 Mark)
In right angled $\triangle PBA$
$PA^2 = PB^2 + (8)^2 = PB^2 + 64$ --- (i) (1/2 Mark)
In right angled $\triangle PAO$
$PA^2 = OP^2 - (10)^2 = (PB + 6)^2 - 100$ --- (ii) (1 Mark)
From (i) and (ii), we have
$PB^2 + 64 = (PB + 6)^2 - 100 \Rightarrow PB = \frac{32}{3}$ cm or $10.6$ cm (1 Mark)