A vertical pole stands at a point O on the horizontal ground. A...

Let the height of the vertical pole be $h$.

At point $A$, the angle of elevation is $60^\circ$. The distance from the base of the pole $O$ to $A$ is:
$OA = h \cot 60^\circ = \dfrac{h}{\sqrt{3}}$

At point $B$, the angle of elevation is $45^\circ$. The distance from the base of the pole $O$ to $B$ is:
$OB = h \cot 45^\circ = h$

Point $A$ is due South of $O$, and point $B$ is due East of $A$. This means the triangle $OAB$ formed on the ground is a right-angled triangle, with the right angle at $A$.

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

Substitute the known values ($AB = 40 \text{ m}$):
\left(\dfrac{h}{\sqrt{3}}\right)^2 + 40^2 = h^2$<br><br>$\dfrac{h^2}{3} + 1600 = h^2$<br><br>$1600 = h^2 - \dfrac{h^2}{3} = \dfrac{2h^2}{3}$<br><br>$h^2 = \dfrac{1600 \times 3}{2} = 2400$<br><br>$h = \sqrt{2400} = 20\sqrt{6} \text{ m}$<br><br>Answer:$20\sqrt{6} \text{ m}