AB is a chord of length 24 cm of a circle of radius 15 cm. The tangents at A and B intersect at a point P. Find the…

AB = $24$ cm
OA = $15$ cm ($\frac{1}{2}$ Mark)
Join OP intersecting AB at Q. Also join OB.
$\triangle$ PAQ $\cong \triangle$ PBQ
$\therefore \angle$ PQA = $\angle$ PQB = $90^{\circ}$ and AQ = QB = $12$ cm (1 Mark)
In right $\triangle$ OQA, OQ = $9$ cm ($\frac{1}{2}$ Mark)
Now, $\triangle$ OAP $\sim \triangle$ OQA ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{OA}{OQ} = \frac{AP}{QA}$ ($\frac{1}{2}$ Mark)
$\Rightarrow AP = \frac{OA \times QA}{OQ} = \frac{15 \times 12}{9} = 20$ cm