Answer :
Let AB be the tower with height of tower AB = h
At C the angle of depression of car measures 30 and 12 minutes later it reaches D where angle of depression is 45.
Let CD = x, D = y
Here, AB = h, ∠ACB = ∠XAC = 30⁰
∠ADB = ∠XAD = 45⁰
In ∆ACB,
x + y = √3h ……(1)
In ∆ABD,
H = y ………(2)
From equation 1 and 2,
x + y = √3y
x = (√3 – 1)y
The distance covered by car in 12 minutes is CD = x
So, time taken to cover distance x is 12 minutes.
So, y = Time taken to cover distance DB
= 6 × 2.73
= 16.38 minutes
The time taken by car to reach the tower is 16.38 minutes.
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