A pole of height 6 m casts a shadow `2sqrt3` m long on the ground. Find the sun's elevation.

Answer

To find the sun's elevation angle when a pole of height 6 m casts a shadow of length \(2\sqrt{3}\) m, we can use trigonometric ratios in a right triangle.

### Step-by-Step Solution:

1. **Identify the Right Triangle**:
- The pole represents the perpendicular side (height) of the triangle, which is 6 m.
- The shadow represents the base of the triangle, which is \(2\sqrt{3}\) m.

2. **Use the Tangent Function**:
- The tangent of the angle of elevation \(\theta\) can be defined as:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\text{Height of the pole}}{\text{Length of the shadow}} \]
- Substituting the values:
\[ \tan(\theta) = \frac{6}{2\sqrt{3}} \]

3. **Simplify the Expression**:
- Simplifying \(\frac{6}{2\sqrt{3}}\):
\[ \tan(\theta) = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

4. **Determine the Angle**:
- Now, we need to find \(\theta\) such that:
\[ \tan(\theta) = \sqrt{3} \]
- From trigonometric values, we know:
\[ \tan(60^\circ) = \sqrt{3} \]
- Therefore, \(\theta = 60^\circ\).

5. **Conclusion**:
- The angle of elevation of the sun is \(60^\circ\).