A pole on the ground leans `60^(@)` to the vertical. At a point a meters away from the base of the pole on the ground, the two halves of the pole subtend the same angle. If the pole and the point are in the same vertical plane, the length of the pole is

Options

A. `3a`
B. `sqrt3a`
C. `(a sqrt3)/(3)`
D. none of these

Answer

To solve the problem step by step, we will analyze the situation involving the leaning pole and the angles formed.

### Step 1: Understand the Geometry
The pole leans at an angle of \(60^\circ\) to the vertical. This means that the angle between the pole and the ground is \(30^\circ\) (since \(90^\circ - 60^\circ = 30^\circ\)).

### Step 2: Define Variables
Let:
- \(L\) = length of the pole
- \(A\) = distance from the base of the pole to point \(A\) on the ground
- \(C\) = midpoint of the pole

Since the pole is leaning, we can visualize two triangles formed by the pole and the point \(A\).

### Step 3: Establish Relationships
The two halves of the pole subtend the same angle at point \(A\). Let’s denote the angle at point \(A\) as \(\alpha\). Since \(C\) is the midpoint of the pole, we have:
- Length of \(BC = CD = \frac{L}{2}\)

### Step 4: Use Trigonometric Ratios
From triangle \(BCA\) (and similarly from triangle \(DCA\)):
- The angle \(BCA = DCA = \alpha\)
- The height from point \(C\) to the ground (which is the vertical height of the pole) is \(h = \frac{L}{2} \cdot \sin(30^\circ) = \frac{L}{2} \cdot \frac{1}{2} = \frac{L}{4}\).
- The horizontal distance from point \(A\) to point \(C\) is \(AC = A\).

Using the tangent function:
\[ \tan(\alpha) = \frac{h}{AC} = \frac{\frac{L}{4}}{A} \]

### Step 5: Set Up the Equation
From the geometry of the problem, we know that:
\[ \tan(60^\circ) = \sqrt{3} \]
Thus, we can write:
\[ \sqrt{3} = \frac{\frac{L}{4}}{A} \]
Rearranging gives us:
\[ L = 4A\sqrt{3} \]

### Step 6: Conclusion
The length of the pole \(L\) is:
\[ L = 4A\sqrt{3} \]