Top of a mountain is observed from A and B at the sea level. If N is the point vertically below P and `angle NAB=alpha, angle NBA=beta, angle NAP=theta, angle NBP=phi`, then `tan phi sin beta= tan theta sin alpha`.

Answer

To solve the problem, we will analyze the angles and the relationships between the points A, B, and P (the top of the mountain) with respect to point N (the point vertically below P).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let \( P \) be the top of the mountain.
- Let \( N \) be the point vertically below \( P \) at sea level.
- Points \( A \) and \( B \) are at sea level from which the angles to the top of the mountain are observed.

2. **Identifying Angles**:
- The angles are defined as follows:
- \( \angle NAB = \alpha \)
- \( \angle NBA = \beta \)
- \( \angle NAP = \theta \)
- \( \angle NBP = \phi \)

3. **Using Right Triangles**:
- In triangle \( ANP \):
- \( \tan(\theta) = \frac{h}{AN} \) where \( h \) is the height of the mountain.
- Therefore, \( AN = \frac{h}{\tan(\theta)} \).

- In triangle \( BNP \):
- \( \tan(\phi) = \frac{h}{BN} \).
- Thus, \( BN = \frac{h}{\tan(\phi)} \).

4. **Using the Law of Sines in Triangle \( ABN \)**:
- From triangle \( ABN \):
- By the sine rule, we have:
\[ \frac{BN}{\sin(\alpha)} = \frac{AN}{\sin(\beta)} \]
- Substituting the expressions for \( AN \) and \( BN \):
\[ \frac{\frac{h}{\tan(\phi)}}{\sin(\alpha)} = \frac{\frac{h}{\tan(\theta)}}{\sin(\beta)} \]

5. **Cross Multiplying**:
- Cross multiplying gives:
\[ h \cdot \sin(\beta) = h \cdot \tan(\phi) \cdot \sin(\alpha) \cdot \tan(\theta) \]
- Canceling \( h \) (assuming \( h \neq 0 \)):
\[ \sin(\beta) = \tan(\phi) \cdot \sin(\alpha) \cdot \tan(\theta) \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ \tan(\phi) \cdot \sin(\beta) = \tan(\theta) \cdot \sin(\alpha) \]

### Final Result:
Thus, we have shown that:
\[ \tan(\phi) \cdot \sin(\beta) = \tan(\theta) \cdot \sin(\alpha) \]