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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.

Answers (1)

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Let the height of the cable tower be $AB = h+7$ m
Given,
The height of the building is $7$ m and the angle of elevation of the top of the tower $\angle A C E=60^{\circ}$, the angle of depression of its foot $\angle B C E=45^{\circ}$.

According to the question,

In triangle $\triangle D B C$,
$\tan 45^{\circ}=\frac{C D}{B D}=\frac{7}{B D}=1$
$B D=7 \mathrm{~m}$
Since $\mathrm{DB}=\mathrm{CE}=7 \mathrm{~m}$
In triangle $\triangle A C E$,
$\tan 60^{\circ}=\frac{h}{C E}=\frac{h}{7}=\sqrt{3}$
$\therefore h=7 \sqrt{3} \mathrm{~m}$
Thus, the total height of the tower is equal to $(7+7\sqrt{3})=7+8.732=15.7 \mathrm{~m}$

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manish

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