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 $$\begin{aligned}& \angle BEC = 30^\circ \\ & therefore \angle ECD = 30^\circ\end{aligned}$$Let CD be $x$ m
$$\begin{aligned}& \frac{DE}{CD} = \tan 30^\circ \Rightarrow \frac{7}{x} = \frac{1}{\sqrt{3}} \Rightarrow x = 7\sqrt{3} \\ & \Rightarrow BE = 7\sqrt{3}\end{aligned}$$Again $$\begin{aligned}& \frac{AB}{BE} = \tan 60^\circ \\ & frac{h-7}{7\sqrt{3}} = \sqrt{3} \\ & h-7 = 7\sqrt{3} \times \sqrt{3} = 7 \times 3 = 21 \\ & h = 28 \\ & therefore\end{aligned}$$ Height of tower = $28$ m