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 use trigonometry and the properties of right triangles.

### Step 1: Define the Variables
Let:
- \( P \) be the top of the pillar.
- \( Q \) be the foot of the pillar.
- \( A \) be the initial point where the man sees the angle of elevation as \( 30^\circ \).
- \( B \) be the point after walking \( 5(\sqrt{3} + 1) \) minutes where the angle of elevation is \( 45^\circ \).
- \( PQ = h \) (the height of the pillar).
- \( QA = d_A \) (the distance from point A to the foot of the pillar).
- \( QB = d_B \) (the distance from point B to the foot of the pillar).

### Step 2: Use Trigonometric Ratios
From point A, the angle of elevation is \( 30^\circ \):
\[ \tan(30^\circ) = \frac{h}{d_A} \implies \frac{1}{\sqrt{3}} = \frac{h}{d_A} \implies d_A = h \sqrt{3} \]

From point B, the angle of elevation is \( 45^\circ \):
\[ \tan(45^\circ) = \frac{h}{d_B} \implies 1 = \frac{h}{d_B} \implies d_B = h \]

### Step 3: Relate Distances
The man walks from point A to point B, covering a distance of \( d_A - d_B \):
\[ d_A - d_B = h \sqrt{3} - h = h(\sqrt{3} - 1) \]

### Step 4: Calculate the Distance from A to B
The distance he covers from A to B in \( 5(\sqrt{3} + 1) \) minutes is:
\[ d_{AB} = h(\sqrt{3} - 1) \]

### Step 5: Calculate Speed
Let \( v \) be the speed of the man:
\[ v = \frac{d_{AB}}{t_{AB}} = \frac{h(\sqrt{3} - 1)}{5(\sqrt{3} + 1)} \]

### Step 6: Calculate Time to Reach from B to Q
To find the time taken to reach from B to Q, we need to use the distance \( d_B = h \) and the speed \( v \):
\[ t_{BQ} = \frac{d_B}{v} = \frac{h}{\frac{h(\sqrt{3} - 1)}{5(\sqrt{3} + 1)}} \]
This simplifies to:
\[ t_{BQ} = \frac{h \cdot 5(\sqrt{3} + 1)}{h(\sqrt{3} - 1)} = \frac{5(\sqrt{3} + 1)}{\sqrt{3} - 1} \]

### Step 7: Rationalize the Denominator
Now we rationalize the denominator:
\[ t_{BQ} = \frac{5(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{5(\sqrt{3} + 1)^2}{3 - 1} = \frac{5(\sqrt{3} + 1)^2}{2} \]
Calculating \( (\sqrt{3} + 1)^2 \):
\[ (\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \]
Thus,
\[ t_{BQ} = \frac{5(4 + 2\sqrt{3})}{2} = 10 + 5\sqrt{3} \]

### Step 8: Substitute the Value of \(\sqrt{3}\)
Substituting \(\sqrt{3} = 1.73\):
\[ t_{BQ} = 10 + 5 \times 1.73 = 10 + 8.65 = 18.65 \text{ minutes} \]

### Final Answer
The time taken by him to reach from point B to the pillar is \( \boxed{18.65} \) minutes.