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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
\(\therefore \) The height of the cable tower is 19.124 m.