Q.1: 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.
Solution:
Answers
Given :-
- 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°
To find :-
- Height of the tower
Solution :-
- Height of tower = AB
- ∠ADB = 30°
- ∠ACB = 60°
- Length of DC = 40m
- DB = DC + CB = 40 + CB
Now, In ∆ACB
→ tan 60° = AB/CB
→ √3 = AB/CB
→ AB = √3CB -----(i)
In ∆ADB
→ tan 30° = AB/DB
Put the value of AB
→ 1/√3 = √3CB/40 + CB
→ 40 + CB = 3CB
→ 3CB - CB = 40
→ 2CB = 40
→ CB = 40/2
→ CB = 20m
Now, height of the tower
From (i)
→ AB = √3CB
→ AB = 20√3
Put the value of √3 is 1.73
→ AB = 20 × 1.73
→ AB = 34.6 m
Hence,
- The height of the tower is 34.6m
Extra Information :-
- tan θ = perpendicular/base
- cot θ = base/perpendicular
- sin θ = perpendicular/hypotenuse
- cosec θ = hypotenuse/perpendicular
- cos θ = base/hypotenuse
- sec θ = hypotenuse/base
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
