A person walking on a straight horizontal road towards a vertical...

Let the height of the tower be $h$ and the final distance from the foot of the tower be $x$.

From the second observation point, the angle of elevation is $30^\circ$:
$\tan 30^\circ = \dfrac{h}{x} \Rightarrow x = h\sqrt{3}$

From the first observation point, the angle of elevation is $15^\circ$ and the distance is $x + 80$:
$\tan 15^\circ = \dfrac{h}{x + 80}$

We know that $\tan 15^\circ = 2 - \sqrt{3}$. Substituting $x = h\sqrt{3}$ into the equation:
$2 - \sqrt{3} = \dfrac{h}{h\sqrt{3} + 80}$

Cross-multiplying gives:
$(2 - \sqrt{3})(h\sqrt{3} + 80) = h$
$2h\sqrt{3} - 3h + 160 - 80\sqrt{3} = h$
$2h\sqrt{3} - 4h = 80\sqrt{3} - 160$
$-2h(2 - \sqrt{3}) = -80(2 - \sqrt{3})$

Dividing both sides by $-2(2 - \sqrt{3})$:
$h = 40$

The height of the tower is $40\text{ m}$.

Answer: $40$