At each end of a horizontal base AB of length 2a the angular height of a certain peak is `15^(@)` and that at the mid-point C of base AB it is `45^(@)`. The height of the peak is

Options

A. `(sqrt(3)-1)/(6)a`
B. `(sqrt(3)-1)/(6)3^(3//4)a`
C. `(3-sqrt(3))/(2)a`
D. `(sqrt(3)-1)/(2sqrt(3))a`

Answer

To find the height of the peak based on the given information, we can use trigonometric principles. Let's denote the height of the peak as \( h \).

1. **Understanding the Geometry**:
- Let \( A \) and \( B \) be the two ends of the horizontal base \( AB \) with length \( 2a \).
- The midpoint \( C \) of \( AB \) is at a distance \( a \) from both \( A \) and \( B \).
- The angular height from points \( A \) and \( B \) to the peak is \( 15^\circ \) and from point \( C \) to the peak is \( 45^\circ \).

2. **Setting Up the Right Triangles**:
- From point \( A \), we can form a right triangle with the height \( h \) and the horizontal distance \( a \).
- From point \( B \), the same right triangle can be formed.
- From point \( C \), the height \( h \) and the horizontal distance \( 0 \) (since it is directly below the peak) can be used.

3. **Using Trigonometric Ratios**:
- From point \( A \):
\[ \tan(15^\circ) = \frac{h}{a} \]
Rearranging gives:
\[ h = a \tan(15^\circ) \tag{1} \]

- From point \( C \):
\[ \tan(45^\circ) = \frac{h}{0} \]
Since \( \tan(45^\circ) = 1 \):
\[ h = a \tag{2} \]

4. **Equating the Heights**:
- From equations (1) and (2), we can equate the two expressions for \( h \):
\[ a \tan(15^\circ) = a \]
- Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ \tan(15^\circ) = 1 \]
- This is incorrect, so we need to analyze the situation again.

5. **Finding the Height Using the Midpoint**:
- Since we have \( h \) from point \( C \) as \( h = a \) and from point \( A \):
\[ h = a \tan(15^\circ) \]
- We can find \( a \) in terms of \( h \):
\[ a = \frac{h}{\tan(15^\circ)} \]
- Substitute \( a \) back into the equation from point \( C \):
\[ h = \frac{h}{\tan(15^\circ)} \]

6. **Solving for \( h \)**:
- Rearranging gives:
\[ h \tan(15^\circ) = h \]
- This implies:
\[ h(1 - \tan(15^\circ)) = 0 \]
- Therefore, the height \( h \) can be calculated using the known value of \( \tan(15^\circ) \).

7. **Final Calculation**:
- Using \( \tan(15^\circ) \approx 0.2679 \):
\[ h = a \tan(15^\circ) \]
- If \( a = 1 \) (for simplicity), then:
\[ h \approx 0.2679 \]

Thus, the height of the peak is approximately \( h \).