Question

The shadow of a tower standing on a level ground found to be 40 m .Longer when the sun's altitude is 30 then when it was 60 . Find the height of the tower.

Answer 1

ʜᴇʀᴇ ɪꜱ ᴛʜᴇ ꜱ_ᴏ_ʟ_ᴜ_ᴛ_ɪ_ᴏ_ɴ

Answer 2

Answer:

Solution :

Let the AB be the height of a tower.

DC = 40 m

DB is 40 m longer than BC.

Therefore,

DB = DC + BC

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In ∆ABC,

tan 60° = AB/BC

√3 = AB/BC (∵ tan 60° = √3)

√3 BC = AB (Taking BC to the left hand side)

BC = AB/√3..........[Equation (1)]

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In ∆ABD,

tan 30° = AB/BD

1/√3 = AB/BD

BD = √3 AB

DC + BC = √3 AB (∵√3 AB DB = DC + BC)

BC = √3 AB - DC

BC = √3 AB - 40 m......[Equation (2)] (∵ It is given that DC = 40 m)

__________________....

★ From Equation (1) and (2) we get,

BC = BC

AB/√3 = √3 AB - 40

AB = √3 (√3 AB) - 40√3 (Taking √3 to the RHS)

AB = 3 AB - 40√3

40√3 = 3AB - AB

40√3 = 2AB

AB = 40√3/2

AB = 20√3 m

or

AB = 20 (1.732) (Taking the value of √3 = 1.732)

AB = 34.64 m