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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.
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.
Solution 1
Let AB be a building and CD be a cable tower.
In ΔABD,
`"AB"/"BD"` = tan 45°
`7/"BD"` = 1
BD = 7 m
In ΔACE,
AE = BD = 7 m
`"CE"/"AE"` = tan 60°
`"CE"/7 = sqrt3`
`"CE" = 7sqrt3 "m"`
CD = CE + ED
= `(7sqrt3 + 7)"m"`
= `7(sqrt3 + 1)"m"`
Therefore, the height of the cable tower is `7(sqrt3+1)"m"`.
Solution 2
Let AB be the 7m high building and CD be the cable tower,
We have,
AB = 7 m, ∠CAE = 60°, ∠DAE = ∠ADB = 45°
Also, DE = AB = 7 m
In ΔABD,
tan 45° = `("AB")/("BD")`
⇒ 1 = `7/("BD")`
⇒ BD = 7 m
So, AE = BD = 7m
Also, In ΔACE,
tan 60° = `("CF")/("AE")`
⇒ `sqrt(3) = ("CE")/7`
⇒ CE = `7sqrt(3)"m"`
Now, CD = CE + DE
= `7 sqrt(3) +7`
= `7 (sqrt(3) +1) "m"`
= 7(1.732 + 1)
= 7(2.732)
= 19.124
= 19.12 m
So, the height of the tower is 19.12 m.
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