A man is walking towards a vertical pillar in a straight path at a uniform speed. At a certain point A on the path, he observes that the angle of elevationof the top of the pillar is `30^(@)`. After walking for `5(sqrt3+1)` minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is `45^(@)`. Then the time taken (in minutes) by him, to reach from B to the pillar, is (take `sqrt3=1.73`)

Answer

To solve the problem step by step, we will analyze the situation involving the man, the pillar, and the angles of elevation.

### Step 1: Understanding the Problem
We have a vertical pillar and a man walking towards it. At point A, the angle of elevation to the top of the pillar is \(30^\circ\). After walking for \(5(\sqrt{3}+1)\) minutes, he reaches point B, where the angle of elevation is \(45^\circ\). We need to find the time taken to reach from point B to the pillar.

### Step 2: Set Up the Scenario
Let:
- \(PQ\) be the height of the pillar (let's denote it as \(H\)).
- \(AQ\) be the distance from point A to the base of the pillar.
- \(BQ\) be the distance from point B to the base of the pillar.

### Step 3: Use Trigonometry at Point B
At point B, the angle of elevation is \(45^\circ\):
\[ \tan(45^\circ) = \frac{H}{BQ} \]
Since \(\tan(45^\circ) = 1\):
\[ 1 = \frac{H}{BQ} \implies BQ = H \]

### Step 4: Use Trigonometry at Point A
At point A, the angle of elevation is \(30^\circ\):
\[ \tan(30^\circ) = \frac{H}{AQ} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{H}{AQ} \implies AQ = H\sqrt{3} \]

### Step 5: Find the Distance AB
The distance \(AB\) can be found as:
\[ AB = AQ - BQ = H\sqrt{3} - H = H(\sqrt{3} - 1) \]

### Step 6: Relate Distance and Time
The man takes \(5(\sqrt{3}+1)\) minutes to walk from A to B. We can denote his speed as \(v\):
\[ v = \frac{AB}{\text{time}} = \frac{H(\sqrt{3} - 1)}{5(\sqrt{3}+1)} \]

### Step 7: Calculate Speed
To find the speed:
\[ v = \frac{H(\sqrt{3} - 1)}{5(\sqrt{3}+1)} \]
Now, we will rationalize the denominator:
\[ v = \frac{H(\sqrt{3} - 1)(\sqrt{3} - 1)}{5((\sqrt{3})^2 - 1^2)} = \frac{H(3 - 2\sqrt{3} + 1)}{5(3 - 1)} = \frac{H(4 - 2\sqrt{3})}{10} = \frac{H(2 - \sqrt{3})}{5} \]

### Step 8: Time from B to Q
The distance from B to Q is \(H\). Using the speed we found:
\[ \text{Time from B to Q} = \frac{\text{Distance}}{\text{Speed}} = \frac{H}{\frac{H(2 - \sqrt{3})}{5}} = \frac{5}{2 - \sqrt{3}} \]

### Step 9: Rationalize the Time
Rationalizing the time:
\[ \text{Time} = \frac{5(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{5(2 + \sqrt{3})}{4 - 3} = 5(2 + \sqrt{3}) = 10 + 5\sqrt{3} \]

### Step 10: Substitute the Value of \(\sqrt{3}\)
Given \(\sqrt{3} \approx 1.73\):
\[ \text{Time} = 10 + 5(1.73) = 10 + 8.65 = 18.65 \text{ minutes} \]

### Final Answer
The time taken by the man to reach from point B to the pillar is approximately **18.65 minutes**.