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 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 boat at Point C is 600, the elevation angle from C to point B will be 600.
Let’s consider the distance covered by the boat from point D to point C is D1 and the distance from C to point A at the foot of the tower is D2.
Step 2: Find the distance D1:
In Δ ABD,
tan 30 = ![]()
∴ ![]()
∴ AD = H√ 3….. (i)
Similarly, in Δ ABC,
tan 60 = ![]()
∴ √ 3 = ![]()
∴ D2 =
….. (ii)
We can see that CD = AD – AC or D1 = AD – D2
Substituting the values from equations (i) and (ii), we get
D1 = H √3 – ![]()
∴ D1 = H(√3 –
)
∴ D1 = H(
)
∴ D1 = ![]()
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 boat will cover distance D2 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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