PA and PB are tangents to the circle and O is the centre of the circle. The radius is 5 cm and PO is 13 cm. If the area of the triangle PAB is M, then the value of $$\sqrt \frac {M}{15}$$ is:

In $$\triangle$$OAP,

OA$$^2$$ + AP$$^2$$ = OP$$^2$$

$$=$$>  5$$^2$$ + AP$$^2$$ = 13$$^2$$

$$=$$>  25 + AP$$^2$$ = 169

$$=$$>  AP$$^2$$ = 144

$$=$$>  AP = 12 cm

Similarly, BP = 12 cm

In $$\triangle$$OAP,

$$\sin x=\frac{OA}{OP}$$

$$\sin x=\frac{5}{13}$$

$$\cos x=\frac{AP}{OP}$$

$$\cos x=\frac{12}{13}$$

In $$\triangle$$ACP,

$$\sin x=\frac{AC}{AP}$$

$$=$$>  $$\frac{5}{13}=\frac{AC}{12}$$

$$=$$>  AC = $$\frac{60}{13}$$

Similarly, BC = $$\frac{60}{13}$$

$$\cos x=\frac{CP}{AP}$$

$$=$$>  $$\frac{12}{13}=\frac{CP}{12}$$

$$=$$>  CP = $$\frac{144}{13}$$

Area of the triangle PAB = M

$$=$$>  $$\frac{1}{2}$$ x AB x CP = M

$$=$$>  $$\frac{1}{2}$$ x (AC+BC) x CP = M

$$=$$>  $$\frac{1}{2}\times\left(\frac{60}{13}+\frac{60}{13}\right)\times\frac{144}{13}$$ = M

$$=$$>  $$\frac{1}{2}\times\frac{120}{13}\times\frac{144}{13}$$ = M

$$=$$>  M = $$\frac{60}{13}\times\frac{144}{13}$$

$$\therefore\ $$ $$\sqrt{\frac{M}{15}}=\sqrt{\frac{60\times144}{169\times15}}=\sqrt{\frac{4\times144}{169}}=\frac{24}{13}$$

Hence, the correct answer is Option B