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From the top of a $7$ m high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $30^\circ$. Determine the height of the tower.
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Sol.
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 $\frac{AB}{BE} = \tan 60^\circ$
$\Rightarrow \frac{h-7}{7\sqrt{3}} = \sqrt{3}$
$h - 7 = 21$
$h = 28$
$\therefore$ Height of tower = $28$ m
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 $\frac{AB}{BE} = \tan 60^\circ$
$\Rightarrow \frac{h-7}{7\sqrt{3}} = \sqrt{3}$
$h - 7 = 21$
$h = 28$
$\therefore$ Height of tower = $28$ m