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

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

The point $A$ is due West of $O$, and the angle of elevation of the top of the pole from $A$ is $30^\circ$. In the right-angled triangle formed by the pole and the ground distance $OA$:
$\tan 30^\circ = \dfrac{h}{OA} \Rightarrow OA = h \cot 30^\circ = h\sqrt{3}$

The point $B$ is due South of $O$, and the angle of elevation from $B$ is $60^\circ$. Similarly:
$\tan 60^\circ = \dfrac{h}{OB} \Rightarrow OB = h \cot 60^\circ = \dfrac{h}{\sqrt{3}}$

Since $A$ is due West and $B$ is due South of $O$, the angle $\angle AOB$ on the horizontal ground is $90^\circ$.

Applying the Pythagorean theorem in the horizontal right-angled triangle $AOB$:
$OA^2 + OB^2 = AB^2$
$(h\sqrt{3})^2 + \left(\dfrac{h}{\sqrt{3}}\right)^2 = (200)^2$
$3h^2 + \dfrac{h^2}{3} = 40000$
$\dfrac{10h^2}{3} = 40000$
$h^2 = 12000$
$h = \sqrt{12000} = 20\sqrt{30}$

The height of the pole is $20\sqrt{30}\text{ m}$.

Answer: $20\sqrt{30}$