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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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M manish


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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