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Q.12 From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60º and the angle of depression of its foot is 45º. Determine the height of the tower.
Answer :


Let AB be the building of height 7 metres and let CD be the cable tower. It is given that the angle of elevation of the top D of the tower observed from A is 60º and the angle of depression of the base C of the tower observed from A is 45º. Then \( ∠ \ EAD \ = \ 60º \) and \( ∠ \ BCA \ = \ 45º \) . Also AB = 7 m



In \( ∆ \ EAD \), we have

\( \frac{DE}{EA} \ = \ tan60º \ => \ \frac{h}{x} \ = \ \sqrt{3} \) \( => \ h \ = \ \sqrt{3} x \)      (1)

In \( ∆ \ ABC \), we have

\( \frac{AB}{BC} \ = \ tan45º \ => \ \frac{7}{x} \ = \ 1 \) \( => \ x \ = \ 7 \)
     (2)

Putting x = 7 in (1), we get

\( h \ = \ 7 \sqrt{3} \ => \ DE \ = \ 7 \sqrt{3} \)m

∴ \( CD \ = \ CE \ + \ ED \ = \ 7 \ + \ 7\sqrt{3} \ = \ 19.124 \)m

∴ The height of the cable tower is 19.124 m.