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 step by step, we will analyze the situation involving the triangle and the projectile motion of the particle.

### Step 1: Understand the Geometry of the Triangle
We have a triangle with base angles of \(30^\circ\) and \(60^\circ\). Let's denote the vertices of the triangle as \(A\), \(B\), and \(C\), where \(C\) is the vertex opposite the base \(AB\). The height from \(C\) to the base \(AB\) will be denoted as \(CD\).

### Step 2: Draw the Perpendicular
From point \(C\), draw a perpendicular \(CD\) to the base \(AB\). Let \(D\) be the foot of the perpendicular. The lengths of the segments on the base can be defined as follows:
- Let \(AD = x\)
- Let \(DB = r - x\) where \(r\) is the length of the base \(AB\)

### Step 3: Use Trigonometric Ratios
Using the angles of the triangle:
- For angle \(C\) (which is \(90^\circ\)), we can use the tangent function for angles \(30^\circ\) and \(60^\circ\):
\[ \tan 30^\circ = \frac{CD}{AD} = \frac{y}{x} \]
\[ \tan 60^\circ = \frac{CD}{DB} = \frac{y}{r - x} \]

### Step 4: Set Up the Equations
From the definitions above, we have:
- \(y = x \tan 30^\circ = \frac{x}{\sqrt{3}}\)
- \(y = (r - x) \tan 60^\circ = (r - x) \sqrt{3}\)

### Step 5: Equate the Two Expressions for \(y\)
Setting the two expressions for \(y\) equal to each other:
\[ \frac{x}{\sqrt{3}} = (r - x) \sqrt{3} \]

### Step 6: Solve for \(x\)
Cross-multiplying gives:
\[ x = (r - x) \cdot 3 \]
\[ x + 3x = 3r \]
\[ 4x = 3r \implies x = \frac{3r}{4} \]

### Step 7: Substitute Back to Find \(y\)
Substituting \(x\) back into the equation for \(y\):
\[ y = \frac{3r}{4\sqrt{3}} \]

### Step 8: Find \(tan \theta\)
Now we can find \(\tan \theta\):
\[ \tan \theta = \frac{y}{x} = \frac{\frac{3r}{4\sqrt{3}}}{\frac{3r}{4}} = \frac{1}{\sqrt{3}} = \tan 30^\circ \]

### Step 9: Use the Trajectory Equation
Using the trajectory equation:
\[ y = x \tan \theta \left(1 - \frac{x}{r}\right) \]
From this, we can derive that:
\[ \tan \theta = \tan 30^\circ + \tan 60^\circ \]

### Step 10: Calculate \(\tan \theta\)
Calculating the sum:
\[ \tan 30^\circ + \tan 60^\circ = \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{1 + 3}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]

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