From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of…

Let AC be $h$ m, $BC = DE = 7$ m, $AB = (h−7)$ m
$\angle AEB = 60^\circ$ and $\angle BEC = 30^\circ$
$\therefore \angle ECD = 30^\circ$
Let CD be $x$ m
$\frac{DE}{CD} = \frac{7}{x} = \tan 30^\circ \Rightarrow x = 7\sqrt{3}$
$\Rightarrow BE = 7\sqrt{3}$
Again $\frac{AB}{BE} = \tan 60^\circ$
$\frac{h-7}{7\sqrt{3}} = \sqrt{3}$
$h - 7 = 21$
$h = 28$
$\therefore \text{Height of tower} = 28$ m