A particle is thrown over a triangle from one end of a horizontal base and after grazing the vertex falls on the other end of the base. If `30^(@) and 60^(@)` be the base angles and `theta` the angle of projection then `tan theta` is

Options

A. `(2)/(sqrt3)`
B. `(4)/(sqrt3)`
C. `(1)/(3)`
D. 3

Answer

To solve the problem, we need to analyze the projectile motion of a particle thrown over a triangle with given angles at the base. The angles of the triangle are 30° and 60°, and we need to find the value of \(\tan \theta\) where \(\theta\) is the angle of projection.

### Step-by-Step Solution:

1. **Identify the Angles**:
- Let \(\alpha = 30^\circ\) and \(\beta = 60^\circ\).
- The triangle has a horizontal base with these angles.

2. **Set Up the Problem**:
- The particle is projected from point A (one end of the base) and grazes the vertex C before landing at point B (the other end of the base).
- Let \(y\) be the height at vertex C and \(x\) be the horizontal distance from A to C.

3. **Use the Tangent Function**:
- For triangle ACO:
\[ \tan \alpha = \frac{y}{x} \quad \text{(1)} \]
- For triangle COB:
\[ \tan \beta = \frac{y}{r - x} \quad \text{(2)} \]
- Here, \(r\) is the total horizontal distance from A to B.

4. **Combine the Equations**:
- From equations (1) and (2):
\[ y = x \tan \alpha \quad \text{and} \quad y = (r - x) \tan \beta \]
- Setting these equal gives:
\[ x \tan \alpha = (r - x) \tan \beta \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ x \tan \alpha + x \tan \beta = r \tan \beta \]
- Factoring out \(x\):
\[ x (\tan \alpha + \tan \beta) = r \tan \beta \]
- Thus:
\[ x = \frac{r \tan \beta}{\tan \alpha + \tan \beta} \quad \text{(3)} \]

6. **Find \(\tan \theta\)**:
- The equation of the trajectory of the projectile is given by:
\[ y = x \tan \theta - \frac{g x^2}{2u^2 \cos^2 \theta} \]
- At the vertex C, the height \(y\) can also be expressed in terms of \(x\) and \(\tan \theta\):
\[ y = x \tan \theta \quad \text{(4)} \]
- Setting (3) and (4) equal gives:
\[ \frac{r \tan \beta}{\tan \alpha + \tan \beta} \tan \theta = \frac{r \tan \beta}{\tan \alpha + \tan \beta} \]

7. **Substituting Values**:
- We know:
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \quad \text{and} \quad \tan 60^\circ = \sqrt{3} \]
- Therefore:
\[ \tan \theta = \tan 30^\circ + \tan 60^\circ = \frac{1}{\sqrt{3}} + \sqrt{3} \]

8. **Final Calculation**:
- Combine the terms:
\[ \tan \theta = \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{1 + 3}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]

### Conclusion:
Thus, the value of \(\tan \theta\) is:
\[ \tan \theta = \frac{4}{\sqrt{3}} \]