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Class 11 Maths Chapter 3: Trigonometric Functions case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks
A giant wheel at a mela in Lucknow has radius 10 m, and its centre is 12 m above the ground. It turns anticlockwise at a steady rate, making one full revolution in 4 minutes. Meera's seat starts at the lowest point of the wheel. When the wheel has turned through an angle θ (in radians) from the start, the height of her seat above the ground is h = 12 − 10 cos θ metres.
Part 1 · 1 mark
When the wheel has made three-quarters of a revolution, the angle θ it has turned through is
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Answer: (b) 3π/2

One revolution is 2π radian, so three-quarters of it is (3/4) × 2π = 3π/2.

Part 2 · 1 mark
The height of Meera's seat when θ = 2π/3 is
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Answer: (c) 17 m

cos (2π/3) = cos (π − π/3) = −cos (π/3) = −1/2, so h = 12 − 10 × (−1/2) = 17 m. Taking cos (2π/3) as 1/2 gives 7 m.

Part 3 · 2 marks
Find the length of the arc Meera's seat travels in the first 50 seconds, and her height at that moment, correct to two decimal places. (Use π = 3.14 and √6 − √2 = 1.035.)
Answer and marking scheme

50 s is 50/240 = 5/24 of a revolution, so θ = (5/24) × 2π = 5π/12 radian (75°). Arc length l = rθ = 10 × 5π/12 = 25π/6 = 25 × 3.14/6 = 13.08 m. cos 75° = cos (45° + 30°) = (√6 − √2)/4 = 1.035/4 = 0.25875, so h = 12 − 10 × 0.25875 = 12 − 2.5875 = 9.41 m (to two decimal places).

  • θ = 5π/12 and arc = 25π/6 ≈ 13.08 m (1)
  • cos 75° = (√6 − √2)/4 and h ≈ 9.41 m (1)
Or
Find the height of Meera's seat when the wheel has turned through 7π/6 radian, and say whether she is then above or below the centre of the wheel. (Use √3 = 1.732.)
Answer and marking scheme

cos (7π/6) = cos (π + π/6) = −cos (π/6) = −√3/2. h = 12 − 10 × (−√3/2) = 12 + 5√3 = 12 + 8.66 = 20.66 m. This is more than 12 m, so she is above the centre.

  • cos (7π/6) = −√3/2 by cos (π + x) = −cos x (1)
  • h = 12 + 5√3 ≈ 20.66 m, above the centre (1)
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