A pole on the ground lenses `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. `sqrt(3)a`
C. `(a+sqrt(3))/3`
D. none of these

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript while providing clear explanations for each step.

### Step-by-Step Solution

1. **Understanding the Problem**:
- We have a pole that makes an angle of \(60^\circ\) with the vertical.
- There is a point \(C\) located \(a\) meters away from the base \(B\) of the pole.
- The two halves of the pole subtend equal angles at point \(C\).

2. **Diagram Representation**:
- Draw a vertical line to represent the pole.
- Label the top of the pole as \(A\) and the base as \(B\).
- The angle \(ABV\) (where \(V\) is the vertical line) is \(60^\circ\).
- Point \(C\) is \(a\) meters away from point \(B\) horizontally.

3. **Identifying the Midpoint**:
- Let \(E\) be the midpoint of the pole \(AB\).
- Since \(E\) is the midpoint, we have \(AE = EB\).

4. **Using Triangle Properties**:
- The triangles \(BEC\) and \(AEC\) are congruent because:
- \(BE = AE\) (both are equal as \(E\) is the midpoint),
- \(\angle ECB = \angle ECA\) (both angles subtended at point \(C\) are equal),
- \(EC\) is common to both triangles.
- Thus, \(\angle CEA = \angle CEB\).

5. **Finding Angles**:
- Since \(\angle CEA + \angle CEB = 180^\circ\) and both angles are equal, we can conclude that:
\[ \angle CEA = \angle CEB = 90^\circ \]
- Therefore, triangle \(BEC\) has angles \(30^\circ\) (from the \(60^\circ\) angle at \(B\)) and \(90^\circ\).

6. **Applying Trigonometric Ratios**:
- In triangle \(BEC\):
- The angle at \(B\) is \(60^\circ\).
- The angle at \(E\) is \(30^\circ\).
- The sine of \(60^\circ\) gives us the relationship:
\[ \sin(60^\circ) = \frac{EB}{BC} \]
- Let \(EB = x\). Then:
\[ \sin(60^\circ) = \frac{x}{a} \]
- Rearranging gives:
\[ x = a \cdot \sin(60^\circ) = a \cdot \frac{\sqrt{3}}{2} \]

7. **Calculating the Length of the Pole**:
- Since \(AB = AE + EB = x + x = 2x\):
\[ AB = 2x = 2 \left(a \cdot \frac{\sqrt{3}}{2}\right) = a\sqrt{3} \]

8. **Final Result**:
- The length of the pole \(AB\) is:
\[ \text{Length of the pole} = a\sqrt{3} \text{ meters} \]