Question

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. (Use `sqrt3` = 1.73)

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer 1

Let AB be the building of height 20 m and BC the transmission tower of height h meter.

∴ AB = 20 m

∴ BC = h m

O is the point of observation. 

Let the angle of elevation of the bottom and top of the tower at the point o is 45° and 60° respectively.

∴ ∠COA = 60° and ∠BOA = 45° 

In ΔOAB,

tan∠BOA = `"AB"/"AO"`

∴ tan 45° = `20/x`

∴ 1 = `20/x`

∴ x = 20 m

∴ AO = 20 m        ...(1)

In ΔOAC,

tan ∠COA = `"AC"/"AO"`

tan 60° = `"AB + BC"/"AO"`

`sqrt3 = (20 + h)/20`

20`sqrt3` = 20 + h

∴ h = 20`sqrt3` − 20

∴ h = 20(`sqrt3` − 1)

Here, `sqrt3` = 1.73              ...(Given)

∴ h = 20(1.73 − 1)

∴ h = 20 × 0.73

∴ h = 14.64 m

Height of tower (h) = 14.64 m.

Answer 2

Let AB be the building of height 20 m and BC the transmission tower of height h meter.

∴ AB = 20 m

∴ BC = h m

O is the point of observation. Let the angle of elevation of the bottom and top of the tower at the point o be 45° and 60°, respectively.

∴ ∠COA = 60° and ∠BOA = 45° 

In ΔOAB,

tan∠BOA = `"AB"/"AO"`

∴ tan 45° = `20/x`

∴ 1 = `20/x`

∴ x = 20 m

∴ AO = 20 m       ...(1)

In ΔOAC,

tan ∠COA = `"AC"/"AO"`

tan 60° = `"AB + BC"/"AO"`

`sqrt3 = (20 + h)/20`

20`sqrt3` = 20 + h

∴ h = 20`sqrt3` − 20

∴ h = 20(`sqrt3` − 1)

h = 20 × 0.732

h = 14.64

Thus, the height of the tower is 14.64 m.