**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.**

**A****ns:**

**Step 1: Diagram for this question:**

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

Given that the angle of depression from point B at Tower top to car at Point C is 30^{0}, hence the angle of elevation from C to point B will be 30^{0}.

Similarly, since the angle of depression from point B at Tower top to car at Point D is 60^{0}, hence the angle of elevation from D to point B will be 60^{0}.

Let’s consider distance covered by car from point D to point C is D_{1} and distance from C to point A at foot of the tower is D_{2}.

**Step 2:Find the distance D _{1}:**

In Δ ABD,

….. (i)

Similarly, in Δ ABC,

….. (ii)

We can see that CD = AD – CD or D_{1} = AD – D_{2}

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

**Step 3: Find the speed of car**

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

We know that:

**Step 4: Find the time taken to cover CA**

Now the car will cover distance with the speed of

We know that:

Substituting the values of distance and speed, we get

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

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