A man on a cliff observes a boat, at an angle of depression `30^(@)` , which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be `60^(@)` . Assuming that the boat sails at a uniform speed, determine.
how much more time it will take to reach the shore ?

Answer

To solve the problem step by step, we will analyze the situation using trigonometry and the properties of right triangles.

### Step 1: Draw the Diagram
First, we need to visualize the situation. Draw a vertical line representing the cliff, and label the top of the cliff as point A. The boat is initially at point B, and after 3 minutes, it moves to point D, directly below point A. The angle of depression from point A to point B is 30 degrees, and from point A to point D is 60 degrees.

### Step 2: Define Variables
Let:
- \( H \) = height of the cliff (AB)
- \( V \) = speed of the boat (in meters per minute)
- \( T \) = time taken to reach the shore from point D

### Step 3: Use Trigonometric Ratios
Using the angle of depression, we can set up the following relationships using the tangent function:

1. From triangle ABC (where angle A is 30 degrees):
\[ \tan(30^\circ) = \frac{H}{AD} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we have:
\[ \frac{1}{\sqrt{3}} = \frac{H}{AD} \implies H = \frac{AD}{\sqrt{3}} \tag{1} \]

2. From triangle CDB (where angle A is 60 degrees):
\[ \tan(60^\circ) = \frac{H}{DB} \]
Since \(\tan(60^\circ) = \sqrt{3}\), we have:
\[ \sqrt{3} = \frac{H}{DB} \implies H = \sqrt{3} \cdot DB \tag{2} \]

### Step 4: Relate Distances
We know that the distance \( AD \) is the distance the boat traveled in 3 minutes, which is:
\[ AD = 3V \]
The distance \( DB \) is the remaining distance to the shore, which we can denote as \( DB \).

### Step 5: Equate Heights
From equations (1) and (2), we can equate the two expressions for \( H \):
\[ \frac{AD}{\sqrt{3}} = \sqrt{3} \cdot DB \]
Substituting \( AD = 3V \):
\[ \frac{3V}{\sqrt{3}} = \sqrt{3} \cdot DB \]
Simplifying gives:
\[ \sqrt{3} \cdot DB = \frac{3V}{\sqrt{3}} \implies DB = \frac{3V}{3} = V \tag{3} \]

### Step 6: Calculate Total Time to Reach Shore
The total distance from point D to the shore is \( DB \). The time \( T \) taken to cover this distance at speed \( V \) is:
\[ T = \frac{DB}{V} = \frac{V}{V} = 1 \text{ minute} \]

### Step 7: Conclusion
Since the boat takes 1 minute to reach the shore from point D, the answer is:
\[ \text{The boat will take 1 minute more to reach the shore.} \]