Question

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.

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°. Determine the height of the tower.

Answer 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"`.

Answer 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.

Answer 3

Let AB be a building, then AB = 7 cm

CD be the height of tower, so

CD = (7 + h) m

BD = AE = x m

In ΔABD,

`tan 45^circ = "AB"/"BD"`

⇒ `1 = 7/x`

⇒ x = 7 cm

In ΔCEA,

`tan 60^circ = "CE"/"AE"`

⇒ `sqrt(3) = h/x`

⇒ `h = xsqrt(3)`

Putting the value of x, we get

`h = 7sqrt(3)`

Now, CD = CE + ED

= `(7 + 7sqrt(3))` m

Hence, height of tower = `7(1 + sqrt(3))` m

= 7(1 + 1.732) m

= 7 × 2.732 m

= 19.124 m

= 19.12 m  ...(Approx)