The angle of elevation of a certain peak when observed from each end of a horizontal base line of length 2a is found to be `theta`. When observed from the mid-point of the base the angle of elevation is `phi`. The height of the peak is

Options

A. `(a)/(sqrt((costheta-cos Psi)))`
B. `(a)/(sqrt((cot^(2)theta-cot^(2)Psi)))`
C. `(a)/(sqrt((cos^(2)theta-cos^(2)Psi)))`
D. None

Answer

To solve the problem, we need to find the height of the peak based on the angles of elevation observed from two points on a horizontal base of length 2a.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the height of the peak be \( h \).
- The horizontal base is of length \( 2a \).
- The two points from which the angles of elevation are observed are at a distance \( a \) from the foot of the peak on either side.

2. **Using Trigonometry for the Angles of Elevation**:
- From the left end of the base (point A), the angle of elevation is \( \theta \).
- From the right end of the base (point B), the angle of elevation is also \( \theta \).
- From the midpoint of the base (point M), the angle of elevation is \( \phi \).

3. **Setting Up the Equations**:
- From point A:
\[ \tan(\theta) = \frac{h}{a} \]
Rearranging gives:
\[ h = a \tan(\theta) \quad \text{(1)} \]

- From point B (which is also at a distance \( a \) from the peak):
\[ \tan(\theta) = \frac{h}{a} \]
This gives the same equation as above:
\[ h = a \tan(\theta) \quad \text{(2)} \]

- From point M (the midpoint, which is at a distance \( a \) from the peak):
\[ \tan(\phi) = \frac{h}{a} \]
Rearranging gives:
\[ h = a \tan(\phi) \quad \text{(3)} \]

4. **Equating the Heights**:
- From equations (1) and (3):
\[ a \tan(\theta) = a \tan(\phi) \]
- Since \( a \) is common and non-zero, we can cancel it out:
\[ \tan(\theta) = \tan(\phi) \]
- This implies that:
\[ h = a \tan(\phi) \]

5. **Final Height Expression**:
- The height of the peak can be expressed as:
\[ h = a \tan(\phi) \]

### Conclusion:
The height of the peak is given by:
\[ h = a \tan(\phi) \]