Consider a triangular plot ABC with sides AB = 7m, BC = 5m and CA =...

The length of the median $BD$ is given by

$BD = \dfrac{1}{2}\sqrt{2\left(B{C}^{2} + B{A}^{2}\right) - {\left(AC\right)}^{2}}$

$⇒ BD = \dfrac{1}{2}\sqrt{2\left(25 + 49\right) - 36}$

$⇒ BD = \dfrac{1}{2}\sqrt{112}$

$⇒ BD = \dfrac{4\sqrt{7}}{2} = 2\sqrt{7}m .$

Now, let $DT$ be the tower of height $hm$ and it subtends $∠ TBD = 30 °$ at the vertex $B .$

Then, in $∆ TDB$ we have $\tan \left(∠ TBD\right) = \dfrac{TD}{BD}$

$⇒ \tan 30 ° = \dfrac{h}{2\sqrt{7}}$

$⇒ \dfrac{1}{\sqrt{3}} = \dfrac{h}{2\sqrt{7}}$

$⇒ h = 2\sqrt{\dfrac{7}{3}}$

$⇒ h = \dfrac{2}{3}\sqrt{21}m .$