At the foot of a mountain the elevation of its summit is `45^(@)`. After ascending 2000 meter towards the mountain upon an incline of `30^(@)`, the elevation changes to `60^(@)`. What is the height of the mountain.

Options

A. `2000(sqrt(3)-1)` meter
B. `1000(sqrt(3)-1)` meter
C. `2000(sqrt(3)+1)` meter
D. `1000(sqrt(3)+1)` meter

Answer

To solve the problem step by step, we will analyze the situation using trigonometry and the given angles of elevation.

### Step 1: Understand the problem
We have a mountain with an angle of elevation of 45 degrees at its base. After ascending 2000 meters towards the mountain at an incline of 30 degrees, the angle of elevation changes to 60 degrees. We need to find the height of the mountain.

### Step 2: Set up the scenario
Let:
- \( h \) = height of the mountain
- \( A \) = point at the foot of the mountain
- \( B \) = point after ascending 2000 meters
- \( C \) = summit of the mountain

### Step 3: Use the first angle of elevation (45 degrees)
From point \( A \), the angle of elevation to the summit \( C \) is 45 degrees. Using the tangent function:
\[ \tan(45^\circ) = \frac{h}{d} \]
where \( d \) is the horizontal distance from point \( A \) to point \( C \).

Since \( \tan(45^\circ) = 1 \):
\[ h = d \quad (1) \]

### Step 4: Use the second angle of elevation (60 degrees)
After ascending 2000 meters towards the mountain at an incline of 30 degrees, we need to find the new horizontal distance from point \( B \) to point \( C \).

The horizontal distance \( d' \) from point \( B \) to point \( C \) can be calculated as follows:
- The vertical rise from point \( B \) due to the incline of 30 degrees is:
\[ \text{Vertical rise} = 2000 \sin(30^\circ) = 2000 \times \frac{1}{2} = 1000 \text{ meters} \]
- The horizontal distance covered from point \( B \) is:
\[ \text{Horizontal distance} = 2000 \cos(30^\circ) = 2000 \times \frac{\sqrt{3}}{2} = 1000\sqrt{3} \text{ meters} \]

Thus, the new horizontal distance \( d' \) from point \( B \) to point \( C \) is:
\[ d' = d - 1000\sqrt{3} \quad (2) \]

Using the angle of elevation of 60 degrees from point \( B \):
\[ \tan(60^\circ) = \frac{h - 1000}{d'} \]
Since \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{h - 1000}{d - 1000\sqrt{3}} \quad (3) \]

### Step 5: Substitute equation (1) into equation (3)
From equation (1), we know \( h = d \). Substituting \( h \) in equation (3):
\[ \sqrt{3} = \frac{d - 1000}{d - 1000\sqrt{3}} \]

### Step 6: Cross-multiply and solve for \( d \)
Cross-multiplying gives:
\[ \sqrt{3}(d - 1000\sqrt{3}) = d - 1000 \]
Expanding and rearranging:
\[ \sqrt{3}d - 3000 = d - 1000 \]
\[ \sqrt{3}d - d = 3000 - 1000 \]
\[ (\sqrt{3} - 1)d = 2000 \]
\[ d = \frac{2000}{\sqrt{3} - 1} \quad (4) \]

### Step 7: Substitute \( d \) back to find \( h \)
Using equation (1):
\[ h = d = \frac{2000}{\sqrt{3} - 1} \]

### Step 8: Rationalize the denominator
To simplify:
\[ h = \frac{2000(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{2000(\sqrt{3} + 1)}{3 - 1} = 1000(\sqrt{3} + 1) \]

### Final Answer
The height of the mountain is:
\[ h = 1000(\sqrt{3} + 1) \text{ meters} \]