A kite if flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground . If the length of the string is `40 sqrt(3)` . find the inclination of the string with the ground

Answer

To find the inclination of the string with the ground when a kite is flying at a height of 60 meters and the length of the string is \(40\sqrt{3}\) meters, we can use trigonometric ratios. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Understand the Problem**:
- The kite is flying at a height of \(60\) meters above the ground.
- The length of the string (hypotenuse) is \(40\sqrt{3}\) meters.
- We need to find the angle of inclination (let's call it \(\theta\)) of the string with the ground.

2. **Draw a Right Triangle**:
- Let point \(O\) be the point on the ground where the string is tied.
- Let point \(P\) be the kite in the air.
- Let point \(Q\) be the point directly below the kite on the ground.
- Thus, triangle \(OPQ\) is formed where:
- \(PQ\) (the height) = \(60\) m (perpendicular)
- \(OP\) (the length of the string) = \(40\sqrt{3}\) m (hypotenuse)
- \(OQ\) (the horizontal distance) is unknown.

3. **Use the Sine Function**:
- The sine of angle \(\theta\) is given by the formula:
\[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PQ}{OP} \]
- Substituting the known values:
\[ \sin \theta = \frac{60}{40\sqrt{3}} \]

4. **Simplify the Expression**:
- Simplifying the right side:
\[ \sin \theta = \frac{60}{40\sqrt{3}} = \frac{60 \div 20}{40\sqrt{3} \div 20} = \frac{3}{2\sqrt{3}} \]
- Rationalizing the denominator:
\[ \sin \theta = \frac{3\sqrt{3}}{2 \cdot 3} = \frac{\sqrt{3}}{2} \]

5. **Find the Angle**:
- We know that \(\sin 60^\circ = \frac{\sqrt{3}}{2}\).
- Therefore, \(\theta = 60^\circ\).

6. **Conclusion**:
- The inclination of the string with the ground is \(60^\circ\).