Two observation points A and B lie on a horizontal plane...

Let the height of the tower be $h$, its top be $T$, and its foot on the horizontal plane be $O$.

In the right-angled triangle $TOA$, the angle of elevation at $A$ is $30^\circ$. The distance from $A$ to the top of the tower $T$ is the hypotenuse $TA$.

$TA = h \csc 30^\circ = 2h$

In the right-angled triangle $TOB$, the angle of elevation at $B$ is $45^\circ$. The distance from $B$ to the top of the tower $T$ is the hypotenuse $TB$.

$TB = h \csc 45^\circ = h\sqrt{2}$

We are given that the segment $AB$ subtends a right angle at the top of the tower $T$, meaning $\angle ATB = 90^\circ$.

In the right-angled spatial triangle $ATB$, applying the Pythagorean theorem:

$AB^2 = TA^2 + TB^2$

Substitute the given values and expressions:

$(60\sqrt{6})^2 = (2h)^2 + (h\sqrt{2})^2$

$21600 = 4h^2 + 2h^2$

$6h^2 = 21600$

$h^2 = 3600$

$h = 60$

The height of the tower is $60$ meters.

Answer: $60$