Let the height of the tower be h. \( \Large \triangle PAO, \tan 60 ^{\circ} = \frac{h}{OA} \) => \( \Large OA = h \cot 60 ^{\circ} \) = \( \Large \frac{h}{\sqrt{3}} \) \( \Large In \triangle PBO, \tan 30 ^{\circ} = \frac{h}{OB} \) => \( \Large OB = \frac{h}{\frac{1}{\sqrt{3}}} = \sqrt{3}h \) => \( \Large AB + AO = \sqrt{3}h \) => \( \Large 20 + \frac{h}{\sqrt{3}} = \sqrt{3}h \) => \( \Large h = \frac{20}{\sqrt{3}-\frac{1}{\sqrt{3}}} = \frac{20\sqrt{3}}{2} = 10\sqrt{3} m \)