A pole stands vertically in the center of a square. When 45 ^ o is...

Let $ABCD$ is a square and $OP$ is the pole.
Now, $∠PQO = {45}^{o}$
Also, $DQ = 30$ meters
$QA = 40$ meters
Side of square $= 30 + 40 = 70$ meters
Now in $∆ POQ$,

$∠Q = {45}^{o}$ and $∠O = {90}^{o}$
$∴ ∠P = {45}^{o}$
$∴ OP = OQ = h$ (let)
Now, let a perpendicular is drawn from $O$ to $AD$ which is $OM$
In $∆ OMQ$,
$QM = DM - QD = \dfrac{70}{2} - 30 = 5$ meters
$OM = \dfrac{AB}{2} = \dfrac{70}{2} = 35$ meters
$∴ OQ = \sqrt{{\left(QM\right)}^{2} + {\left(OM\right)}^{2}} = \sqrt{{5}^{2} + {35}^{2}}$
$= 5\sqrt{1 + {7}^{2}} = 25\sqrt{2}$ meters