A man standing at a point P is watching the top of elevation of `30^@`. The man walks some distance towards the tower and then his angle of elevation of thetop of the toweris `60^@`. If the height of the tower 30 m, then the distance he moves is

Options

A. 20 m
B. 20 V 3 m
C. 22 m
D. 22 V3 m

Answer

To solve the problem step by step, we will use trigonometric principles related to angles of elevation.

### Step 1: Understand the Problem
A man is observing the top of a tower from point P with an angle of elevation of \(30^\circ\). He then walks towards the tower and the angle of elevation becomes \(60^\circ\). The height of the tower is given as 30 m. We need to find the distance he moved towards the tower.

### Step 2: Set Up the Diagram
1. Let the height of the tower be \(h = 30\) m.
2. Let the distance from point P to the base of the tower be \(d_1\) when the angle of elevation is \(30^\circ\).
3. Let the distance from the new position (let's call it Q) to the base of the tower be \(d_2\) when the angle of elevation is \(60^\circ\).

### Step 3: Use Trigonometric Ratios
Using the tangent function, we can relate the height of the tower to the distances and angles of elevation.

1. From point P (angle \(30^\circ\)):
\[ \tan(30^\circ) = \frac{h}{d_1} \]
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), so:
\[ \frac{1}{\sqrt{3}} = \frac{30}{d_1} \]
Rearranging gives:
\[ d_1 = 30\sqrt{3} \]

2. From point Q (angle \(60^\circ\)):
\[ \tan(60^\circ) = \frac{h}{d_2} \]
We know that \(\tan(60^\circ) = \sqrt{3}\), so:
\[ \sqrt{3} = \frac{30}{d_2} \]
Rearranging gives:
\[ d_2 = \frac{30}{\sqrt{3}} = 10\sqrt{3} \]

### Step 4: Calculate the Distance Moved
The distance the man moved towards the tower is the difference between \(d_1\) and \(d_2\):
\[ \text{Distance moved} = d_1 - d_2 = 30\sqrt{3} - 10\sqrt{3} = 20\sqrt{3} \]

### Final Answer
The distance the man moved towards the tower is \(20\sqrt{3}\) meters.

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