Let $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is a square and $\mathrm{O}\mathrm{P}$ is the pole.
Now, $\angle \mathrm{P}\mathrm{Q}\mathrm{O}={45}^{\mathrm{o}}$
Also, $\mathrm{D}\mathrm{Q}=30$ meters
$\mathrm{Q}\mathrm{A}=40$ meters
Side of square $=30+40=70$ meters
Now in $∆\mathrm{P}\mathrm{O}\mathrm{Q}$,

$\angle \mathrm{Q}={45}^{\mathrm{o}}$ and $\angle \mathrm{O}={90}^{\mathrm{o}}$
$\therefore \angle \mathrm{P}={45}^{\mathrm{o}}$
$\therefore \mathrm{O}\mathrm{P}=\mathrm{O}\mathrm{Q}=h$ (let)
Now, let a perpendicular is drawn from $\mathrm{O}$ to $\mathrm{A}\mathrm{D}$ which is $\mathrm{O}\mathrm{M}$
In $∆\mathrm{O}\mathrm{M}\mathrm{Q}$,
$\mathrm{Q}\mathrm{M}=\mathrm{D}\mathrm{M}-\mathrm{Q}\mathrm{D}=\frac{70}{2}-30=5$ meters
$OM=\frac{AB}{2}=\frac{70}{2}=35$ meters
$\therefore OQ=\sqrt{{\left(\mathrm{Q}\mathrm{M}\right)}^2+{\left(\mathrm{O}\mathrm{M}\right)}^2}=\sqrt{5^2+{35}^2}$
$=5\sqrt{1+7^2}=25\sqrt{2}$ meters