In a `DeltaABC` whose cicumradius is 2 . A pole standing inside the `DeltaABC` and angle of elevation of top of the pole from points A,B,C is `60^@` then find height of pole

Answer

To find the height of the pole standing inside triangle ABC, we can follow these steps:

### Step 1: Understand the given information
We are given:
- The circumradius \( R \) of triangle ABC is 2.
- The angle of elevation from points A, B, and C to the top of the pole is \( 60^\circ \).

### Step 2: Set up the problem
Let:
- \( P \) be the top of the pole.
- \( Q \) be the base of the pole on the plane of triangle ABC.
- The height of the pole is \( h \).
- The distance from point \( Q \) to each vertex (A, B, C) is equal to the circumradius \( R = 2 \).

### Step 3: Use trigonometry to express height
From point A, we can use the tangent of the angle of elevation to express the height of the pole:
\[ \tan(60^\circ) = \frac{h}{AQ} \]
Where \( AQ \) is the horizontal distance from point A to the base of the pole \( Q \).

### Step 4: Calculate \( AQ \)
Since the circumradius \( R \) is the distance from the center of the circumcircle to any vertex, and \( AQ \) is the horizontal distance from point A to the base of the pole, we know:
\[ AQ = R = 2 \]

### Step 5: Substitute into the tangent equation
Now substituting \( AQ \) into the tangent equation:
\[ \tan(60^\circ) = \sqrt{3} \]
Thus, we have:
\[ \sqrt{3} = \frac{h}{2} \]

### Step 6: Solve for \( h \)
Now, we can solve for the height \( h \):
\[ h = 2 \cdot \sqrt{3} \]

### Step 7: Calculate the numerical value
Calculating \( h \):
\[ h = 2 \cdot \sqrt{3} \approx 2 \cdot 1.732 = 3.464 \]

### Final Answer
The height of the pole is approximately \( 3.464 \).

---