A contractor plans to install two slides for the children to play in a park. For the children below the age of 6 years, he prefers to have a slide whose top is at a height of 2·0 m and is inclined at an angle of 30° to the ground, whereas for older children, he wants to have a steep slide at a height of 4·0 m and inclined at an angle of 60° to the ground. What would be the length of the slide in each case ?

MathsCBSE Class 10 · Basic · Board exam 2024, set 430/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2024: question 33

Set in the CBSE Class 10 board exam of 2024, paper 430/4/1, section D, question 33, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry

A contractor plans to install two slides for the children to play in a park. For the children below the age of 6 years, he prefers to have a slide whose top is at a height of $2 \cdot 0$ m and is inclined at an angle of $3 0^{\circ}$ to the ground, whereas for older children, he wants to have a steep slide at a height of $4 \cdot 0$ m and inclined at an angle of $6 0^{\circ}$ to the ground. What would be the length of the slide in each case ?

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/4/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

The slide is the hypotenuse and the height is the side opposite the given angle.

Hint 2

Use sin θ = height/slide length with θ = 30° and θ = 60°.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Figure: two right triangles. In $△ A B C$ (right-angled at B), BC = 2 m is the height, AC is the slide and $∠ A = 3 0^{\circ}$ . In $△ P QR$ (right-angled at Q), QR = 4 m is the height, PR is the slide and $∠ P = 6 0^{\circ}$ .
1 mark
In $△ A B C$ : $sin 3 0^{\circ} = \frac{B C}{A C}$ .
1 mark
$\frac{2}{A C} = \frac{1}{2}$
½ mark
$\Rightarrow A C = 4$ m. So the slide for children below 6 years is 4 m long.
½ mark
In $△ P QR$ : $sin 6 0^{\circ} = \frac{QR}{P R}$ .
1 mark
$\frac{4}{P R} = \frac{3}{2}$
½ mark
$\Rightarrow P R = \frac{8}{3} = \frac{8 3}{3} \approx 4.62$ m. So the slide for older children is about 4.62 m long.
½ mark

Answer: 4 m for children below 6 years; 8√3/3 m ≈ 4.62 m for older children

Where the marks are

Correct labelled figures (1 mark)
sin 30° = BC/AC (1 mark)
Substitution 2/AC = 1/2 (½ mark)
AC = 4 m (½ mark)
sin 60° = QR/PR (1 mark)
Substitution 4/PR = √3/2 (½ mark)
PR = 8√3/3 ≈ 4.62 m (½ mark)

Revise the chapter

Chapter: Some Applications of Trigonometry (notes, formulas, where marks are lost and every past question)
Topic: Heights and Distances: One Right Triangle

Practise this chapterBoard-style questions from Some Applications of Trigonometry, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 430/4/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2024: question 33

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry