Question 4 of 12beginner🔧 ApplyNumerical3 marks

A kite is 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. 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.

Correct Answer

The length of the string is 403\sqrt{3} m (or approximately 69.28 m).

Let the position of the kite be point A and the point where the string is tied to the ground be point C. The height of the kite from the ground is AB = 60 m.

In the right triangle ABC:

  • AB = 60 m (perpendicular)
  • Angle of inclination ∠ACB = 60°
  • AC = length of string (hypotenuse)

Using the trigonometric ratio: sin θ = perpendicular/hypotenuse

sin 60° = AB/AC

3\sqrt{3}/2 = 60/AC

Therefore, AC = 60×260 \times 2/3\sqrt{3} = 120/3\sqrt{3}

Rationalizing the denominator: AC = 120/3\sqrt{3} × 3\sqrt{3}/3\sqrt{3} = 1203\sqrt{3}/3 = 403\sqrt{3} m

Hence, the length of the string is 403\sqrt{3} m.

Exercise: EXERCISE 9.1 | Q: 5 | (Page Page 9)
For More Understanding

Explanation

This question from the textbook (Question 5) tests the application of trigonometric ratios in real-life situations. Similar to Example 2 in the context, where an electrician needs to find the length of a ladder using sin 60°, this problem requires finding the hypotenuse (string length) when the perpendicular (height) and angle are known. The sin ratio is chosen because it relates the perpendicular to the hypotenuse, which matches the given values.

Solution Steps

  1. Step 1: Draw a right triangle ABC where A is the kite position, B is the point directly below the kite on ground, and C is where the string is tied.

  2. Step 2: Identify AB = 60 m (height/perpendicular) and ∠ACB = 60° (angle of inclination).

  3. Step 3: Apply sin 60° = perpendicular/hypotenuse = AB/AC

  4. Step 4: Substitute values: 3\sqrt{3}/2 = 60/AC

  5. Step 5: Solve for AC: AC = 60×260 \times 2/3\sqrt{3} = 120/3\sqrt{3} = 403\sqrt{3} m

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