$A$ tower stands vertically on the ground. From a point on the ground,which is $15\, m$ away from the foot of the tower,the angle of elevation of the top of the tower is found to be $60^{\circ}.$ Find the height of the tower. (in $m$)

$(15\sqrt{3})$ First,let us represent the problem with a right-angled triangle $ABC$ (as shown in the figure),where $AB$ is the height of the tower,$BC = 15\, m$ is the distance of the point from the foot of the tower,and $\angle ACB = 60^{\circ}$ is the angle of elevation.
In the right-angled triangle $ABC$,we have:
$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}$
$\tan 60^{\circ} = \frac{AB}{BC}$
Since $\tan 60^{\circ} = \sqrt{3}$,we substitute the values:
$\sqrt{3} = \frac{AB}{15}$
$AB = 15\sqrt{3}\, m$
Thus,the height of the tower is $15\sqrt{3}\, m$.