The angles of depression of the top and the bottom of a 50 m high building from the top of a tower are 45° and 60° ,…

Correct figure
Let AB be the tower of height H m and CD be the building.
In $\triangle ABD$,
$\tan 60^\circ = \sqrt{3} = \frac{H}{X}$
$\Rightarrow H = \sqrt{3} X$ --------(i)
In $\triangle AEC$
$\tan 45^\circ = 1 = \frac{H-50}{X}$
$\Rightarrow X = H-50$ --------(ii)
Solving equations (i) & (ii)
$H = 25(3 + \sqrt{3}) = 25(3 + 1.73) = 118.25$
The height of tower is $118.25$ m