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 step by step, we will use trigonometric ratios to find the height of the mount.

### Step 1: Understand the problem
We have a tower that is 50 m high, standing on top of a mount of unknown height \( h \). The angles of elevation from a point on the ground to the top and bottom of the tower are \( 75^\circ \) and \( 60^\circ \) respectively.

### Step 2: Set up the diagram
Let:
- \( A \) be the top of the tower,
- \( B \) be the bottom of the tower,
- \( C \) be the point on the ground directly below the tower,
- \( D \) be the point from where the angles of elevation are measured.

The height of the tower \( AB = 50 \) m, and the height of the mount \( BC = h \).

### Step 3: Identify the triangles
From point \( D \):
- The angle of elevation to point \( A \) (top of the tower) is \( 75^\circ \).
- The angle of elevation to point \( B \) (bottom of the tower) is \( 60^\circ \).

### Step 4: Use the tangent function
Using the tangent function for triangle \( DBC \):
\[ \tan(60^\circ) = \frac{BC}{DC} = \frac{h}{DC} \]
From trigonometric tables, we know that \( \tan(60^\circ) = \sqrt{3} \). Therefore:
\[ \sqrt{3} = \frac{h}{DC} \implies DC = \frac{h}{\sqrt{3}} \tag{1} \]

Now, using triangle \( DAC \):
\[ \tan(75^\circ) = \frac{AC}{DC} = \frac{50 + h}{DC} \]
From trigonometric tables, we can use the identity for \( \tan(75^\circ) \):
\[ \tan(75^\circ) = \tan(45^\circ + 30^\circ) = \frac{\tan(45^\circ) + \tan(30^\circ)}{1 - \tan(45^\circ) \tan(30^\circ)} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

### Step 5: Substitute DC from (1) into the equation
Substituting \( DC = \frac{h}{\sqrt{3}} \) into the equation for \( \tan(75^\circ) \):
\[ \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{50 + h}{\frac{h}{\sqrt{3}}} \]
Cross-multiplying gives:
\[ (\sqrt{3} + 1) \cdot \frac{h}{\sqrt{3}} = (50 + h)(\sqrt{3} - 1) \]

### Step 6: Solve for h
Expanding both sides:
\[ h(\sqrt{3} + 1) = (50 + h)(\sqrt{3} - 1) \cdot \sqrt{3} \]
This simplifies to:
\[ h(\sqrt{3} + 1) = 50(\sqrt{3} - 1) + h(\sqrt{3} - 1) \]
Rearranging gives:
\[ h(\sqrt{3} + 1 - \sqrt{3} + 1) = 50(\sqrt{3} - 1) \]
Thus:
\[ h(2) = 50(\sqrt{3} - 1) \]
Finally, we find:
\[ h = 25(\sqrt{3} - 1) \]

### Step 7: Conclusion
The height of the mount \( h \) is \( 25(\sqrt{3} - 1) \) meters.