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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.
Solution 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.
Solution 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.
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