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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