A vertical tower stands at the center of a circular park. Three...

The center of the circular park is the circumcenter of the horizontal triangle $ABC$. The distance from any point on the boundary (like $A$) to the center is the circumradius $R$.

In $\Delta ABC$, $AB = AC = 40$ and $BC = 48$. Draw the altitude $AD$ to $BC$. Since $ABC$ is isosceles, $D$ is the midpoint of $BC$, so $BD = 24$.
By Pythagoras theorem in $\Delta ABD$:
$AD = \sqrt{AB^2 - BD^2} = \sqrt{40^2 - 24^2} = 32$

Let $O$ be the circumcenter, which lies on $AD$. Let $OA = R$, then $OD = 32 - R$.
In right-angled $\Delta OBD$:
$OB^2 = OD^2 + BD^2$
$R^2 = (32 - R)^2 + 24^2$
$R^2 = 1024 - 64R + R^2 + 576$
$64R = 1600$
$R = 25$

The angle of elevation of the top of the tower from $A$ is $\theta = \tan^{-1}(2)$, so $\tan \theta = 2$.
The height of the tower $h$ is given by:
$h = R \tan \theta = 25 \times 2 = 50$

Answer: $50$