A tower PQ stands on a horizontal ground with base Q on the ground....

Plotting the diagram of the given values we get,

From the right triangle $RQS,$

$\tan 60 ° = \dfrac{15}{a}$

$a = \dfrac{15}{\sqrt{3}} = 5\sqrt{3}$

In right triangle $PQS,$

$\tan 75 ° = \dfrac{h}{a}$

$⇒ \dfrac{h}{5\sqrt{3}} = 2 + \sqrt{3}$

$⇒ h = 10\sqrt{3} + 15$

$⇒ h = 5\left(2\sqrt{3} + 3\right)$