Q) A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30∘, which is approaching to the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60∘. Find the further time taken by the car to reach the foot of the tower.

Ans:

Step 1: Diagram for this question:

A straight highway leads to the foot of a tower. A man standing at the top of the tower

Here, let’s consider AB is the tower and the car initially at point D, and then after 6 seconds, it is at point C.

Given that the angle of depression from point B at the Tower top to the car at Point D is 300, the elevation angle from D to point B will be 300.

Similarly, since the angle of depression from point B at the Tower top to the car at Point C is 600, the elevation angle from C to point B will be 600.

Let’s consider the distance covered by car from point D to point C is D1 and distance from C to point A at foot of the tower is D2.

Step 2:Find the distance D1:

In Δ ABD,

\tan 30 = \frac{AB}{AD}

\therefore \frac{1}{\sqrt3} = \frac{H}{AD}

\therefore AD = H\sqrt3….. (i)

Similarly, in Δ ABC,

\tan 60 = \frac{AB}{AC}

\therefore \sqrt3 = \frac{H}{D_2}

\therefore D_2 = \frac{H}{\sqrt3} ….. (ii)

We can see that CD = AD – CD or D1 = AD – D2

Substituting the values from equations (i) and (ii), we get

D_1 = H\sqrt3 - \frac{H}{\sqrt3}

\therefore D_1 = H(\sqrt3 - \frac{1}{\sqrt3})

\therefore D_1 = H(\frac{3-1}{\sqrt3})

\therefore D_1 = \frac{2H}{\sqrt3}

Step 3: Find the speed of car

Now, it is given that this distance CD is covered in 6 secs

We know that: Speed = \frac{Distance}{Time}

\therefore Speed = \frac {\frac{2H}{\sqrt3}}{6}

\therefore Speed = \frac{H}{3\sqrt3}

Step 4: Find the time taken to cover CA

Now the car will cover distance D_2 with the speed of \frac{H}{3\sqrt3}

We know that: Speed = \frac{Distance}{Time}

\therefore Time = \frac{Distance}{Speed}

Substituting the values of distance and speed, we get

Time = \frac{(\frac{H}{\sqrt3})}{(\frac{H}{3\sqrt3})}

\therefore Time = \frac{(\frac{\cancel {H}}{\cancel{\sqrt3}})}{(\frac{\cancel{H}}{3\cancel{\sqrt3}})}

\therefore Time = 3

Therefore, the distance from point C to point A will be covered in 3 secs.

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