A particle is thrown over a triangle from one end of horizontal base and grazing over the vertex falls on the other end of the base. If `alpha, beta` are the base angles and `theta` the angle of projection, then

Options

A. `tantheta= tanalpha-tanbeta`
B. `tantheta= tanbeta-tanalpha`
C. `tantheta= tanalpha+ tanbeta`
D. None of these

Answer

To solve the problem, we need to analyze the projectile motion of a particle thrown over a triangle. The triangle has a horizontal base with angles α and β at the ends, and the particle grazes the vertex of the triangle before landing on the other end of the base.

### Step-by-Step Solution:

1. **Understand the Geometry**:
- Let the triangle have a horizontal base with points A and B at the ends, and vertex C at the top. The angles at points A and B are α and β respectively.

2. **Identify the Projectile Motion**:
- The particle is projected from point A with an initial velocity \( u \) at an angle \( \theta \) with the horizontal. It grazes point C (the vertex) and lands at point B.

3. **Define the Coordinates**:
- Let the horizontal distance from A to B be \( R \) (the range). The height at point C (the vertex) can be denoted as \( h \).

4. **Use the Tangent Function**:
- The height \( h \) can be expressed in terms of the angles:
\[ h = R \tan \alpha \quad \text{(for angle α)} \]
\[ h = R \tan \beta \quad \text{(for angle β)} \]
- Since both expressions equal \( h \), we can set them equal to each other:
\[ R \tan \alpha = R \tan \beta \]

5. **Relate the Angles**:
- The particle grazes the vertex C, which means that the total vertical distance covered by the projectile when it reaches point C must equal the height of the triangle:
\[ h = R \tan \theta \]

6. **Combine the Equations**:
- From the previous steps, we have:
\[ R \tan \theta = R \tan \alpha + R \tan \beta \]
- Dividing through by \( R \) (assuming \( R \neq 0 \)):
\[ \tan \theta = \tan \alpha + \tan \beta \]

7. **Conclusion**:
- The relationship between the angles of projection and the base angles of the triangle is given by:
\[ \tan \theta = \tan \alpha + \tan \beta \]

### Final Answer:
Thus, the relationship derived from the problem is:
\[ \tan \theta = \tan \alpha + \tan \beta \]