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Question

The string of a kite is 100 meters long, and it makes an angle of 30° with the horizontal. Find the height of the kite from the ground.

The correct answer is

50 m

Understanding the Kite Height Problem

This problem involves finding the height of a kite using trigonometry. We are given the length of the kite string and the angle the string makes with the horizontal ground. This scenario forms a right-angled triangle where:

  • The kite string is the hypotenuse (the longest side, opposite the right angle).
  • The height of the kite from the ground is the side opposite the angle given (the vertical side).
  • The horizontal distance from the person flying the kite to the point directly below the kite is the adjacent side.

We are given the length of the hypotenuse (string length) and the angle, and we need to find the length of the opposite side (height).

Identifying the Relevant Trigonometric Ratio

In a right-angled triangle, the trigonometric ratios relate the angles to the sides. The ratio that connects the opposite side and the hypotenuse is the sine function.

The sine of an angle ($\theta$) in a right-angled triangle is defined as:

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

In our problem:

  • \(\theta = 30^\circ\) (the angle the string makes with the horizontal)
  • Hypotenuse = 100 meters (the length of the kite string)
  • Opposite Side = Height of the kite (what we need to find)

Setting up the Calculation

We can rearrange the sine formula to solve for the opposite side (height):

\(\text{Height} = \text{Hypotenuse} \times \text{sin}(\theta)\)

Substituting the given values:

\(\text{Height} = 100 \text{ m} \times \text{sin}(30^\circ)\)

Calculating the Height of the Kite

To find the height, we need the value of \(\text{sin}(30^\circ)\). The sine of 30 degrees is a standard trigonometric value, which is 0.5 or \(\frac{1}{2}\).

\(\text{sin}(30^\circ) = 0.5\)

Now, substitute this value into the equation:

\(\text{Height} = 100 \text{ m} \times 0.5\)

\(\text{Height} = 50 \text{ m}\)

So, the height of the kite from the ground is 50 meters.

Summary of Given Information and Result

Quantity Value
Length of kite string (Hypotenuse) 100 m
Angle with horizontal (\(\theta\)) \(30^\circ\)
Trigonometric Ratio Used Sine (\(\text{sin}(\theta)\))
Calculated Height (Opposite Side) 50 m

Conclusion

By applying the sine function, which relates the angle of elevation to the opposite side (height) and the hypotenuse (string length) in a right-angled triangle, we calculated the height of the kite. The height of the kite is 50 meters.

This problem demonstrates a practical application of basic trigonometry in everyday scenarios.

Revision Table: Trigonometry Basics

Term Definition Relevance to Problem
Hypotenuse The side opposite the right angle in a right triangle. Kite string length (100 m)
Opposite Side The side opposite the angle of interest. Height of the kite
Adjacent Side The side next to the angle of interest (not the hypotenuse). Horizontal distance (not needed for this problem)
Sine (\(\text{sin}(\theta)\)) Ratio of the length of the opposite side to the length of the hypotenuse. Used to find the height: \(\text{sin}(30^\circ) = \frac{\text{Height}}{\text{String Length}}\)

Additional Information: Standard Trigonometric Values

It is useful to remember the sine, cosine, and tangent values for common angles like 0°, 30°, 45°, 60°, and 90°.

  • \(\text{sin}(0^\circ) = 0\), \(\text{cos}(0^\circ) = 1\), \(\text{tan}(0^\circ) = 0\)
  • \(\text{sin}(30^\circ) = \frac{1}{2}\), \(\text{cos}(30^\circ) = \frac{\sqrt{3}}{2}\), \(\text{tan}(30^\circ) = \frac{1}{\sqrt{3}}\)
  • \(\text{sin}(45^\circ) = \frac{1}{\sqrt{2}}\), \(\text{cos}(45^\circ) = \frac{1}{\sqrt{2}}\), \(\text{tan}(45^\circ) = 1\)
  • \(\text{sin}(60^\circ) = \frac{\sqrt{3}}{2}\), \(\text{cos}(60^\circ) = \frac{1}{2}\), \(\text{tan}(60^\circ) = \sqrt{3}\)
  • \(\text{sin}(90^\circ) = 1\), \(\text{cos}(90^\circ) = 0\), \(\text{tan}(90^\circ)\) is undefined

Knowing these values makes solving trigonometry problems quicker and easier.

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Important Questions from Geometry

  1. A rectangle has a length of 24 cm and diagonals of length 25 cm each. The area of the rectangle (in cm²) is:

  2. In which quadrants do the points (-2, 3) and (3, -2) lie?

  3. If in ΔABC, AB = 5 cm, BC = 12 cm, and AC = 13 cm, then the length of the median BE is:

  4. In a △ABC right-angled at B, AB = 8 units and AC = 10 units. What is the value of sin2θ−cos2θ where θ is ∠ACB?

  5. The length of the side of an equilateral triangle is 43​ cm. Find its height:

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