The shadow of a tower standing on a level ground is found to be 40 m longer when the sun's altitude is 30° than when it was 60° . Find the height of the tower.
Answers
Answer:-
Let AB be the tower and BC be the length of its shadow when sun’s altitude is 60° and DB be the length of the shadow when the angle of elevation is 30°.
Let us assume:-
AB = h m
BC = x m
DB = (40 + x) m
In right triangle ABC,
tan 60° = AB/BC
√3 = h/x
h = √3 x → (i)
In right triangle ABD,
tan 30° = AB/BD
1/√3 =h/(x + 40) → (ii)
From (i) and (ii),
x(√3 )(√3 ) = x + 40
3x = x + 40
2x = 40
x = 20
Substituting x = 20 in (i), we get:-
h = 20√3
Therefore, the height of the tower is 20√3 m.
Note:- Look into the attechment for the diagram.
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
