A kite is flying at a height of 60 m above the ground. The string arrached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is `60^(@)`. Find the length of the string, assuming that there is no slack in the string.

Options

A. `20sqrt3`
B. `40sqrt3`
C. `60sqrt3`
D. `80sqrt3`

Explanation

To find the length of the string tied to the kite flying at a height of 60 m with an inclination of 60 degrees to the ground, we can use trigonometric ratios. Here’s the step-by-step solution:

### Step 1: Understand the Problem
We have a right triangle formed by the height of the kite, the ground, and the string. The height of the kite (perpendicular) is 60 m, and the angle of elevation (inclination of the string with the ground) is 60 degrees.

### Step 2: Identify the Triangle Components
In the right triangle:
- The height of the kite (perpendicular) = 60 m
- The angle of elevation = 60 degrees
- The length of the string (hypotenuse) = ?

### Step 3: Use the Sine Function
We can use the sine function, which relates the angle to the opposite side (height) and the hypotenuse (length of the string).

\[ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \]

Here, \(\theta = 60^\circ\), the opposite side is the height of the kite (60 m), and the hypotenuse is the length of the string (let's call it \(L\)).

\[ \sin(60^\circ) = \frac{60}{L} \]

### Step 4: Substitute the Value of Sine
The value of \(\sin(60^\circ)\) is \(\frac{\sqrt{3}}{2}\).

\[ \frac{\sqrt{3}}{2} = \frac{60}{L} \]

### Step 5: Solve for the Length of the String
Cross-multiply to solve for \(L\):

\[ \sqrt{3} \cdot L = 120 \]

Now, divide both sides by \(\sqrt{3}\):

\[ L = \frac{120}{\sqrt{3}} \]

### Step 6: Rationalize the Denominator
To simplify \(\frac{120}{\sqrt{3}}\), multiply the numerator and denominator by \(\sqrt{3}\):

\[ L = \frac{120 \sqrt{3}}{3} = 40 \sqrt{3} \]

### Final Answer
The length of the string is \(40\sqrt{3}\) meters.

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