**Q) A man on a cliff observes a boat at an angle of depression of 30° which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be 60°. Find the time taken by the boat from here to reach the shore.**

**Ans: **Let’s start with the diagram for this question:

Here, let’s consider AB is the cliff and the boat initially at point D, and then after 6 minutes, it is at point C.

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

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

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

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

In Δ ABD,

tan 30 =

∴

∴ AD = H√ 3….. (i)

Similarly, in Δ ABC,

tan 60 =

∴ √ 3 =

∴ D_{2} = ….. (ii)

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

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

D_{1} = H √3 –

∴ D_{1} = H(√3 – )

∴ D_{1} = H()

∴ D_{1 }=

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

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

We know that: Speed =

∴ Speed =

∴ Speed =

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

Now the car will cover distance D_{2} with the speed of

We know that: Speed =

∴ Time =

Substituting the values of distance and speed, we get

Time =

∴ Time =

∴ Time = 3 minutes

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

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