Consider a triangular plot A B C with sides A B = 7   m , B C = 5   m and C A = 6   m . A…

The length of the median $BD$ is given by 

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

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

$\Rightarrow BD=\frac{1}{2}\sqrt{112}$

$\Rightarrow BD=\frac{4\sqrt{7}}{2}=2\sqrt{7} m.$

Now, let $DT$ be the tower of height $h m$ and it subtends $\angle TBD=30°$ at the vertex $B.$

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

$\Rightarrow \tan 30°=\frac{h}{2\sqrt{7}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{2\sqrt{7}}$

$\Rightarrow h=2\sqrt{\frac{7}{3}}$

$\Rightarrow h=\frac{2}{3}\sqrt{21} m.$