A man from the top of a 100 m high tower sees a car moving towards the tower at an angle of depression of 30°. after some time, the angle of depression becomes 60°. the distance ( in metres) travelled by the car during this time is
Answers
The distance ( in metres) travelled by car during this time is 115.47 m.
Step-by-step explanation:
Referring to the figure attached below, let’s make some assumptions,
AB = height of the tower = 100 m
The first angle of depression, ∠EAC = ∠ACB = 30°
As the car moves towards the tower, the second angle of depression becomes, ∠EAD = ∠ADB = 60°
In ∆ ACB, applying the trigonometry ratios of a triangle, we get
tan 30° = perpendicular/base = AB/CB
⇒ tan 30° = 100/CB
⇒ 1/√3 = 100/CB
⇒ CB = 100√3 m …… (i)
In ∆ ADB, applying the trigonometry ratios of a triangle, we get
tan 60° = perpendicular/base = AB/DB
⇒ tan 60° = 100/DB
⇒ √3 = 100/DB
⇒ DB = 100/√3 m …… (ii)
Thus,
The distance (in metres) travelled by car during this time i.e., distance covered during travelling from point C to D(as shown in the figure) is given by,
= CD
= CB – DB
= 100√3 – (100/√3)
= 173.20 – 57.73
= 115.47 m
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