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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° and the angle of depression of its foot is 45°. Determine the height of the tower.

12. 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 depression of its foot is 45°. Determine the height of the tower.

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Let the height of the cable tower be (AB = $h+7$)m
Given,
Height of the building is 7 m and angle of elevation of the top of the tower $\angle ACE = 60^o$ , angle of depression of its foot $\angle BCE = 45^o$.

According to question,

In triangle $\Delta DBC$,
$\\\tan 45^o = \frac{CD}{BD} = \frac{7}{BD} = 1\\ BD =7\ m$
since DB = CE = 7 m

In triangle $\Delta ACE$,

$\\\tan 60^o = \frac{h}{CE}=\frac{h}{7}=\sqrt{3}\\ \therefore h = 7\sqrt{3}\ m$

Thus, the total height of the tower equal to $h+7$ $=7(1+\sqrt{3}\) m$

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