The thread of a kite makes angle `60^(@)` with the horizontal plane . If the length of the thread be 80 m , then the vertical height of the kite will be

Options

A. `(40)/(sqrt(3))` metre
B. `30sqrt(3)` metre
C. 80 metre
D. `40sqrt(3)` metre

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Identify the Triangle
We have a right triangle formed by the thread of the kite, the vertical height of the kite, and the horizontal distance from the point directly below the kite to the point where the thread is attached. The angle between the thread and the horizontal plane is given as \(60^\circ\).

### Step 2: Use Trigonometric Ratios
In this triangle:
- The length of the thread is the hypotenuse (80 m).
- The vertical height of the kite is the opposite side (perpendicular).
- The horizontal distance is the adjacent side (base).

### Step 3: Calculate the Horizontal Distance (Base)
Using the cosine function:
\[ \cos(60^\circ) = \frac{\text{Base}}{\text{Hypotenuse}} \]
Substituting the known values:
\[ \cos(60^\circ) = \frac{\text{Base}}{80} \]
We know that \(\cos(60^\circ) = \frac{1}{2}\):
\[ \frac{1}{2} = \frac{\text{Base}}{80} \]
Now, solving for the base:
\[ \text{Base} = 80 \times \frac{1}{2} = 40 \text{ m} \]

### Step 4: Calculate the Vertical Height (Perpendicular)
Using the tangent function:
\[ \tan(60^\circ) = \frac{\text{Perpendicular}}{\text{Base}} \]
Substituting the known values:
\[ \tan(60^\circ) = \frac{\text{Perpendicular}}{40} \]
We know that \(\tan(60^\circ) = \sqrt{3}\):
\[ \sqrt{3} = \frac{\text{Perpendicular}}{40} \]
Now, solving for the perpendicular (vertical height):
\[ \text{Perpendicular} = 40 \times \sqrt{3} = 40\sqrt{3} \text{ m} \]

### Final Answer
The vertical height of the kite is \(40\sqrt{3}\) meters.

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