Question

A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from each corner of the park be π/3. If the radius of the circumcircle of ΔABC is 2, then the height of the pole is equal to :

• A. 2$\sqrt{3}$ / 3
• B. 2$\sqrt{3}$
• C. 1 / $\sqrt{3}$
• D. $\sqrt{3}$

Hint:

If the angle of elevation is same from all vertices, the foot of the pole is the circumcenter. If same from all sides, it is the incenter.

Answer 1

The Correct Option is B

To solve this problem, we need to determine the height of the pole given the conditions specified.

Given that the angle of elevation of the top of the pole from each corner of the triangular park ABC is \(\frac{\pi}{3}\) and the radius of the circumcircle of \(\Delta ABC\) is 2, we must use trigonometric concepts to find the height of the pole.

Step-by-Step Solution:

1. The circumradius (R) of \(\Delta ABC\) is given as 2.
2. We assume that the length of each side of the triangle (s) touches the circumcircle, hence let O be the center of the circumcircle.
3. The angle of elevation from each corner means the pole's height forms an angle \(\frac{\pi}{3}\) with the horizontal plane from each vertex.
4. Let \(h\) be the height of the pole.
5. Using trigonometry, for each vertex, we use the tangent relation:

The formula for tangent from a vertex equals the height divided by the horizontal distance to that height (which is the circumradius here since vertices are points on the circumcircle):

\[ \tan\left(\frac{\pi}{3}\right) = \frac{h}{R} \]

Given \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\) and \(R = 2\), substitute these into the equation:

\[ \sqrt{3} = \frac{h}{2} \]

Solving for \(h\), we get:

\[ h = 2 \times \sqrt{3} \]

Thus, the height of the pole is indeed \(2\sqrt{3}\).

Conclusion:

The correct answer is 2\sqrt{3}, which confirms that the height of the pole is \(2\sqrt{3}\).