A man watches a boat travelling towards him while standing on a top. Its deviation is `30^o` . After 20sec deviation changes to `45^o`. Then how much time it talkes to reach bottom of the tower ?

Answer

To solve the problem, we will use trigonometry and the relationships between the angles and distances involved.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the height of the tower be \( h \).
- The man is observing the boat at two different angles: \( 30^\circ \) and \( 45^\circ \).
- Let the distance of the boat from the base of the tower at the first observation (when the angle is \( 30^\circ \)) be \( x \) and at the second observation (when the angle is \( 45^\circ \)) be \( x + y \), where \( y \) is the distance the boat has traveled towards the tower in 20 seconds.

2. **Using the Angle of Deviation**:
- From the triangle formed by the height of the tower and the distance to the boat, we can write:
- For \( 30^\circ \):
\[ \tan(30^\circ) = \frac{h}{x} \implies \frac{1}{\sqrt{3}} = \frac{h}{x} \implies h = \frac{x}{\sqrt{3}} \]
- For \( 45^\circ \):
\[ \tan(45^\circ) = \frac{h}{x+y} \implies 1 = \frac{h}{x+y} \implies h = x + y \]

3. **Equating the Two Expressions for \( h \)**:
- From the two equations for \( h \):
\[ \frac{x}{\sqrt{3}} = x + y \]
- Rearranging gives:
\[ y = \frac{x}{\sqrt{3}} - x = x\left(\frac{1}{\sqrt{3}} - 1\right) = x\left(\frac{1 - \sqrt{3}}{\sqrt{3}}\right) \]

4. **Finding the Speed of the Boat**:
- Since the boat travels distance \( y \) in 20 seconds, we have:
\[ y = s \cdot 20 \implies s = \frac{y}{20} \]

5. **Substituting for \( y \)**:
- Substitute \( y \) into the speed equation:
\[ s = \frac{x\left(\frac{1 - \sqrt{3}}{\sqrt{3}}\right)}{20} \]

6. **Finding the Time to Reach the Bottom**:
- The distance \( x \) can be expressed in terms of speed and time:
\[ x = s \cdot t \]
- Substitute \( s \) into this equation:
\[ x = \frac{x\left(\frac{1 - \sqrt{3}}{\sqrt{3}}\right)}{20} \cdot t \]
- Rearranging gives:
\[ t = \frac{20}{\frac{1 - \sqrt{3}}{\sqrt{3}}} = 20 \cdot \frac{\sqrt{3}}{1 - \sqrt{3}} \]
- Rationalizing:
\[ t = 20 \cdot \frac{\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = 20 \cdot \frac{\sqrt{3}(1 + \sqrt{3})}{-2} = -10\sqrt{3}(1 + \sqrt{3}) \]

7. **Final Calculation**:
- This gives us the time it takes for the boat to reach the bottom of the tower.