A vertical tower stands on a horizontal plane. A car is driving...

Let the height of the tower be $h$ and its base be $O$.

The maximum angle of elevation occurs when the car is at the point of closest approach to the tower. Let this point be $C$. The line $OC$ is perpendicular to the road.

The horizontal distances from the base $O$ to the points $A$, $B$, and $C$ are:
$OA = h \cot 30^\circ = h\sqrt{3}$
$OB = h \cot 45^\circ = h$
$OC = h \cot 60^\circ = \dfrac{h}{\sqrt{3}}$

In the horizontal plane, $\triangle OCA$ is a right-angled triangle at $C$. The distance $AC$ along the road is:
$AC = \sqrt{OA^2 - OC^2} = \sqrt{3h^2 - \dfrac{h^2}{3}} = \sqrt{\dfrac{8}{3}}h$

Similarly, $\triangle OCB$ is right-angled at $C$. The distance $BC$ is:
$BC = \sqrt{OB^2 - OC^2} = \sqrt{h^2 - \dfrac{h^2}{3}} = \sqrt{\dfrac{2}{3}}h$

Notice that $AC = \sqrt{\dfrac{8}{3}}h = 2\sqrt{\dfrac{2}{3}}h = 2BC$.

Since the car travels from $A$ to $B$ and then to $C$, the distance $AB = AC - BC = 2BC - BC = BC$.

Because the distances $AB$ and $BC$ are equal and the car travels at a uniform speed, the time taken to travel from $B$ to $C$ is exactly the same as the time taken to travel from $A$ to $B$.

Therefore, the further time taken is $10$ minutes.

Answer: $10$