Maths · Relations and functions · chapter 2 of 13
Class 12 Maths Chapter 2: Inverse Trigonometric Functions case study question with answers
One case-based question of the kind the board asks: 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 surveyor in Jaipur studies a water tower 30 m tall standing on level ground. She stands x metres from the foot of the tower and measures θ, the angle of elevation of its top, in radians. Since tan θ = 30/x, she records θ = tan⁻¹(30/x). Because x > 0, the angle θ always lies between 0 and π/2, inside the principal value branches of tan⁻¹, cot⁻¹, sin⁻¹ and cos⁻¹, so the same angle can be written with any of these inverse functions.
Part 1 · 1 mark
When she stands 30√3 m from the foot of the tower, θ equals
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Answer: (a) π/6
θ = tan⁻¹(30/(30√3)) = tan⁻¹(1/√3) = π/6, which lies in (−π/2, π/2).
Part 2 · 1 mark
Written with cot⁻¹, θ = cot⁻¹(k), where k is
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Answer: (c) x/30
cot θ = 1/tan θ = x/30, and θ ∈ (0, π/2) ⊂ (0, π), so θ = cot⁻¹(x/30).
Part 3 · 2 marks
She finds that θ = sin⁻¹(3/5). Find her distance x from the foot of the tower.
Answer and marking scheme
sin θ = 3/5 with θ ∈ (0, π/2), so cos θ = √(1 − 9/25) = 4/5 and tan θ = 3/4. Then 30/x = 3/4, so x = 40 m.
- tan θ = 3/4 from sin θ = 3/5 with θ acute (1)
- x = 40 m (1)
Or
Write θ in the form cos⁻¹(…) in terms of x, and find x when θ = cos⁻¹(12/13).
Answer and marking scheme
The line of sight has length √(x2 + 900), so cos θ = x/√(x2 + 900) and, θ being in [0, π], θ = cos⁻¹(x/√(x2 + 900)). If cos θ = 12/13, sin θ = 5/13 and tan θ = 5/12, so 30/x = 5/12 and x = 72 m.
- θ = cos⁻¹(x/√(x^{2} + 900)) (1)
- x = 72 m (1)