A vertical tower stands at a point O on horizontal ground. A point...

Let the height of the tower be $h$ meters.

Since the angle of elevation of the top of the tower from $A$ is $45^{\circ}$, the distance $OA = h \cot 45^{\circ} = h$.

Since the angle of elevation from $B$ is $30^{\circ}$, the distance $OB = h \cot 30^{\circ} = h\sqrt{3}$.

Point $A$ is due East of $O$ and point $B$ is due North of $O$, so $\triangle AOB$ is a right-angled triangle at $O$.

Using the Pythagorean theorem in $\triangle AOB$:
$AB^2 = OA^2 + OB^2$

Substitute the given values:
$(20)^2 = h^2 + (h\sqrt{3})^2$
$400 = h^2 + 3h^2$
$400 = 4h^2$
$h^2 = 100$

Therefore, $h = 10$ meters.

Answer: $10$