From a point of the ground, angle of elevation of top of a tower is `30^(@)`. On moving `50sqrt(3)` meter towards the tower the elevation changes to `60^(@)`. What is the height of the tower ?

Options

A. `75sqrt(3)` m
B. 75 m
C. `90sqrt(3)` m
D. `60sqrt(3)` m

Answer

To solve the problem step by step, we will use trigonometric principles related to right triangles.

### Step 1: Define the variables
Let:
- \( AB \) be the height of the tower.
- \( BC = x \) be the distance from the point of observation to the base of the tower before moving.
- \( CD = 50\sqrt{3} \) be the distance moved towards the tower.
- \( BD = BC + CD = x + 50\sqrt{3} \) be the distance from the new point of observation to the base of the tower.

### Step 2: Set up the first triangle (before moving)
From the point of observation at \( C \):
- The angle of elevation to the top of the tower \( A \) is \( 30^\circ \).
- Using the tangent function:
\[ \tan(30^\circ) = \frac{AB}{BC} \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so we can write:
\[ \frac{1}{\sqrt{3}} = \frac{AB}{x} \]
Rearranging gives us:
\[ AB = \frac{x}{\sqrt{3}} \quad \text{(Equation 1)} \]

### Step 3: Set up the second triangle (after moving)
From the new point of observation at \( D \):
- The angle of elevation to the top of the tower \( A \) is \( 60^\circ \).
- Using the tangent function again:
\[ \tan(60^\circ) = \frac{AB}{BD} \]
We know that \( \tan(60^\circ) = \sqrt{3} \), so we can write:
\[ \sqrt{3} = \frac{AB}{x + 50\sqrt{3}} \]
Rearranging gives us:
\[ AB = \sqrt{3}(x + 50\sqrt{3}) \quad \text{(Equation 2)} \]

### Step 4: Equate the two expressions for \( AB \)
From Equation 1 and Equation 2, we have:
\[ \frac{x}{\sqrt{3}} = \sqrt{3}(x + 50\sqrt{3}) \]

### Step 5: Solve for \( x \)
Multiplying both sides by \( \sqrt{3} \) to eliminate the fraction:
\[ x = 3(x + 50\sqrt{3}) \]
Expanding the right side:
\[ x = 3x + 150\sqrt{3} \]
Rearranging gives:
\[ x - 3x = 150\sqrt{3} \]
\[ -2x = 150\sqrt{3} \]
Dividing by -2:
\[ x = -75\sqrt{3} \]
Since distance cannot be negative, we take the positive value:
\[ x = 75\sqrt{3} \]

### Step 6: Substitute \( x \) back to find \( AB \)
Now substitute \( x \) back into Equation 1 to find \( AB \):
\[ AB = \frac{75\sqrt{3}}{\sqrt{3}} = 75 \]

### Conclusion
The height of the tower \( AB \) is \( 75 \) meters.