From the top of a $70\, m$ tall tower,the angle of depression of the top and the base of a pole are found to be $30^{\circ}$ and $45^{\circ}$ respectively. Find the distance of the pole from the tower and the height of the pole.

(N/A) Let the height of the tower be $AB = 70\, m$ and the height of the pole be $CD = h$. Let the distance between the tower and the pole be $x$.
In $\triangle ABD$,$\tan(45^{\circ}) = \frac{AB}{BD} \implies 1 = \frac{70}{x} \implies x = 70\, m$.
Now,consider the top of the pole $C$. Draw a horizontal line from $C$ meeting $AB$ at $E$. Then $AE = h$ and $EB = 70 - h$.
In $\triangle AEC$,$\tan(30^{\circ}) = \frac{AE}{EC} \implies \frac{1}{\sqrt{3}} = \frac{70 - h}{70}$.
$70 = \sqrt{3}(70 - h) \implies 70 = 70\sqrt{3} - h\sqrt{3} \implies h\sqrt{3} = 70(\sqrt{3} - 1)$.
$h = \frac{70(\sqrt{3} - 1)}{\sqrt{3}} = 70(1 - \frac{1}{\sqrt{3}}) \approx 70(1 - 0.577) = 70(0.423) = 29.61\, m$.
The distance is $70\, m$ and the height is approximately $29.61\, m$.