Question

From the top of a 7m 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.

Answer 1

Correct Answer - `19.12 m`
Let AB be the building and CD be the cable tower.
Draw `BE_|_CD` . Let `CD = h` metres.
Then, `CE = AB = 7 m` and `DE = (h-7) m`.
From right `DeltaCAB` , we have
`(AC)/(AB) = cot 45^(@) = 1 rArr (AC)/(7m) = 1`.
`rArr AC = 7`.
`:. BE = AC = 7 m`.
From right `DeltaBED` , we have
`(DE)/(BE) = tan 60^(@) = sqrt(3) rArr (h-7)/(7) = sqrt(3)`.
`:. h =7sqrt(3) + 7 = 7 (sqrt(3) + 1) = 7(1.732 +1)`
`rArr h =(7xx 2.732) = 19.12`.