the shadow of a tower standing on a level ground found to be 40√3 m .longer when the sun's altitude is 30 ° then when it was 60° . find the height of the tower.
Answers
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Correct Question :The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
Step-by-step explanation:
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)
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★ 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
