The angles of elevation of a vertical tower
standing inside a triangular field at the vertices of the field
are each equal to $\theta$. If the length of the sides of the field are
30 m, 50 m and 70 m, the height of the tower is

Let $OP$ be the tower of height $h$at the point
$O$ in the triangular field $ABC$ with sides $BC = 30, CA = 50$

and $AB = 70$.

Then  $OA=OB=OC=h\cot \theta$

$\Rightarrow O$ is the circumcentre of the triangle $ABC$

$\Rightarrow h\cot \theta =R$ (the radius of the circumcircle)

$\Rightarrow h=R\tan \theta$

We know that $R=abc/4\mathrm{\Delta }$

where   $\mathrm{\Delta }=\sqrt{s(s-a)(s-b)(s-c)}$

             $=\sqrt{75\times 45\times 25\times 5}=25\times 5\times 3\sqrt{3}$

Thus,      $h=\frac{30\times 50\times 70}{4\times 25\times 5\times 3\sqrt{3}}\tan \theta$

                  $=\frac{70}{\sqrt{3}}\tan \theta \mathrm{m}.$