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, value of $$\sqrt {\frac{{\text{M}}}{{15}}} $$ is :

Solution (By Examveda Team)

$$\eqalign{ & PA = 12{\text{ cm}} \cr & \sin \theta = \frac{{AC}}{{AP}} = \frac{{AO}}{{OP}} \cr & \frac{{AC}}{{12}} = \frac{5}{{13}} \cr & AC = \frac{{60}}{{13}} \cr & \tan \theta = \frac{{AC}}{{PC}} = \frac{{AO}}{{AP}} \cr & \frac{{\frac{{60}}{{13}}}}{{PC}} = \frac{5}{{12}} \cr & PC = \frac{{144}}{{13}} \cr & {\text{Area of }}\Delta PCB = \frac{1}{2} \times PC \times AC \cr & {\text{Area of }}\Delta PAB = PC \times AC \cr & M = \frac{{144}}{{13}} \times \frac{{60}}{{13}} \cr & \frac{{\sqrt M }}{{15}} = \sqrt {\frac{{144 \times 60}}{{13 \times 13 \times 15}}} = \frac{{24}}{{13}} \cr} $$