A tower 50 m high , stands on top of a mount, from a point on the ground the angles of elevation of the top and bottom of the tower are found to be `75^@` and `60^@` respectively. The height of the mount is

Options

A. 25 m
B. `25(sqrt3-1)`m
C. `25sqrt3`m
D. `25(sqrt3+1)` m

Answer

To solve the problem, we need to find the height of the mount (h) given the height of the tower (50 m) and the angles of elevation from a point on the ground to the top and bottom of the tower.

### Step-by-Step Solution:

1. **Understand the Geometry**:
- Let the height of the mount be \( h \).
- The total height from the ground to the top of the tower is \( h + 50 \) meters.
- The angles of elevation from point P on the ground to the top of the tower (point A) and the bottom of the tower (point B) are \( 75^\circ \) and \( 60^\circ \), respectively.

2. **Set Up the Right Triangles**:
- For the bottom of the tower (point B), we can use the angle of elevation \( 60^\circ \):
\[ \tan(60^\circ) = \frac{h}{d} \]
where \( d \) is the horizontal distance from point P to the base of the mount.

3. **Calculate \( d \)**:
- We know \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{h}{d} \implies d = \frac{h}{\sqrt{3}} \]

4. **For the Top of the Tower (point A)**:
- Using the angle of elevation \( 75^\circ \):
\[ \tan(75^\circ) = \frac{h + 50}{d} \]

5. **Substitute \( d \)**:
- Substitute \( d \) from step 3 into the equation:
\[ \tan(75^\circ) = \frac{h + 50}{\frac{h}{\sqrt{3}}} \]
- This simplifies to:
\[ \tan(75^\circ) = \frac{(h + 50) \sqrt{3}}{h} \]

6. **Calculate \( \tan(75^\circ) \)**:
- We can express \( \tan(75^\circ) \) using the angle addition formula:
\[ \tan(75^\circ) = \tan(45^\circ + 30^\circ) = \frac{\tan(45^\circ) + \tan(30^\circ)}{1 - \tan(45^\circ) \tan(30^\circ)} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

7. **Cross-Multiply**:
- Rearranging gives:
\[ h \tan(75^\circ) = (h + 50) \sqrt{3} \]
- Expanding this:
\[ h \tan(75^\circ) = h \sqrt{3} + 50 \sqrt{3} \]

8. **Collect Terms**:
- Rearranging gives:
\[ h (\tan(75^\circ) - \sqrt{3}) = 50 \sqrt{3} \]
- Thus:
\[ h = \frac{50 \sqrt{3}}{\tan(75^\circ) - \sqrt{3}} \]

9. **Calculate \( h \)**:
- Substitute \( \tan(75^\circ) \) and calculate \( h \).

### Final Calculation:
Using the approximate value \( \tan(75^\circ) \approx 3.732 \):
\[ h = \frac{50 \cdot 1.732}{3.732 - 1.732} \approx \frac{86.6}{2} \approx 43.3 \text{ meters} \]

### Final Answer:
The height of the mount is approximately \( 43.3 \) meters.