The shadow of a tower standing on level ground is found to be 40 m longer when Sun’s altitude is 30° than when it was 60°. Find the height of the tower.
Answers
When the shadow of a tower standing on level ground is found to be 40 m longer when Sun’s altitude is 30° than when it was 60°, the height of the tower is 3.46m.
1. Let the height of the tower be h meters and the length of shadow formed at 60° altitude of sun be x meters.
2. The diagram corresponding to the shawdow formed is given below.
3. From the figure we find that
tan 30° = h/(x+4)
=>1/root(3) = h/(x+4)
=> x = rt(3)*h- 4 ---equation 1
tan 60° = h/x
=>rt(3)*x= h
Substituting equation in the above equation
=>rt(3)*(rt(3)*h- 4)= h
=> 3h -4rt(3) =h
=>h= 2rt(3) =3.46m
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
________________
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)]
________________
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
