A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is `30^(@)` (Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is `45^(@)`. Then the time taken (in seconds) by the boat from B to reach the base of the tower is:

Options

A. 10
B. `10sqrt(3)`
C. `10(sqrt(3)+1)`
D. `10(sqrt(3)-1)`

Answer

To solve the problem, we will use trigonometry and the concept of relative motion. Let's break it down step by step.

### Step 1: Understand the Geometry
Let the height of the tower be \( h \). The man is observing the boat from the top of the tower. At point A, the angle of depression to the boat is \( 30^\circ \). After 20 seconds, the angle of depression changes to \( 45^\circ \) at point B.

### Step 2: Set Up the Relationships
From the angle of depression, we can set up the following relationships using trigonometric ratios:

1. At point A (angle \( 30^\circ \)):
\[ \tan(30^\circ) = \frac{h}{x} \]
where \( x \) is the horizontal distance from the base of the tower to point A.
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x} \implies h = \frac{x}{\sqrt{3}} \tag{1} \]

2. At point B (angle \( 45^\circ \)):
\[ \tan(45^\circ) = \frac{h}{y} \]
where \( y \) is the horizontal distance from the base of the tower to point B.
Since \( \tan(45^\circ) = 1 \), we have:
\[ 1 = \frac{h}{y} \implies h = y \tag{2} \]

### Step 3: Relate \( x \) and \( y \)
From equations (1) and (2), we can equate the two expressions for \( h \):
\[ \frac{x}{\sqrt{3}} = y \implies y = \frac{x}{\sqrt{3}} \tag{3} \]

### Step 4: Determine the Distance Covered by the Boat
The boat travels from point A to point B in 20 seconds. The distance covered by the boat is:
\[ AB = x - y \]
Substituting equation (3) into this:
\[ AB = x - \frac{x}{\sqrt{3}} = x\left(1 - \frac{1}{\sqrt{3}}\right) = x\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) \]
Let the speed of the boat be \( s \). The distance covered in 20 seconds is:
\[ AB = s \cdot 20 \implies s \cdot 20 = x\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) \tag{4} \]

### Step 5: Find the Time to Reach the Base of the Tower
Now, we need to find the time taken by the boat to travel from point B to the base of the tower (point C). The distance BC is:
\[ BC = y = h \tag{2} \]
Using equation (2) where \( h = y \):
\[ y = h = \frac{x}{\sqrt{3}} \]
The time \( t \) taken to travel from B to C is given by:
\[ t = \frac{BC}{s} = \frac{y}{s} = \frac{\frac{x}{\sqrt{3}}}{s} \]
Now, substituting \( s \) from equation (4):
\[ s = \frac{x\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)}{20} \]
Thus,
\[ t = \frac{\frac{x}{\sqrt{3}}}{\frac{x\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)}{20}} = \frac{20}{\sqrt{3}-1} \]

### Step 6: Rationalize the Expression
To rationalize:
\[ t = \frac{20}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{20(\sqrt{3}+1)}{3-1} = \frac{20(\sqrt{3}+1)}{2} = 10(\sqrt{3}+1) \]

### Final Answer
Thus, the time taken by the boat from point B to reach the base of the tower is:
\[ t = 10(\sqrt{3}+1) \text{ seconds} \]