Class 10 Maths Chapter 9 Extra Questions (HOTS Level)
These higher-order-thinking questions go beyond the standard Exercise 9.1 pattern — general proofs, reverse-engineering, assertion-reason, and multi-step word problems, all with numbers not used in the NCERT textbook exercise.
- Q1 (Reverse-engineering). The angle of elevation of a tower’s top from a point 28√3 m from its base is such that its tangent equals 1/√3. Find the tower’s height.
tanθ = 1/√3 means θ = 30°. Height = 28√3 × tan30° = 28√3 × 1/√3 = 28 m. - Q2 (General/abstract proof). From two points at distances p and q (p > q) from a tower’s base, on the same side and in line with it, the angles of elevation are complementary. Prove that the tower’s height is √(pq).
Proof: tanθ = h/q, tan(90°-θ) = cotθ = h/p. Multiplying: tanθ × cotθ = 1 = h²/(pq), so h² = pq, giving h = √(pq). (This generalises the standard textbook result, where p=9, q=4, giving h=6.) - Q3 (Multi-number word problem). A 1.6 m tall boy notices the top of a 21.6 m pillar at 30° elevation, then walks towards it until the elevation is 45°. Find the distance he walked.
Height above eye level = 21.6-1.6 = 20 m. d1 = 20/tan30° = 20√3. d2 = 20/tan45° = 20. Distance walked = 20√3-20 = 20(√3-1) m (approx 14.64 m). - Q4 (Assertion-Reason). Assertion (A): As the sun’s angle of elevation decreases through the afternoon, the shadow of a vertical pole gets longer.
Reason (R): For a pole of fixed height h and sun elevation θ, shadow length = h/tanθ, and tanθ is an increasing function of θ on (0°, 90°).
Options: (a) Both A and R true, R correctly explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Answer: (a). Since tanθ increases with θ, a smaller θ gives a smaller tanθ, hence a larger shadow length h/tanθ — R correctly explains A. - Q5 (Reciprocal-substitution speed/time, fresh numbers). From the top of a 90 m tower, a car’s angle of depression changes from 30° to 60° in 10 seconds as it approaches at constant speed. Find the further time it takes to reach the tower’s foot.
d1 = 90√3, d2 = 90/√3 = 30√3. Distance covered in 10s = 60√3, speed = 6√3/s. Remaining distance d2 = 30√3, time = 30√3/6√3 = 5 seconds. - Q6 (General two-point formula, proof). A tower’s angle of elevation from a point is θ. After moving a distance ‘a’ towards the tower, the angle becomes φ (φ > θ). Prove the tower’s height is h = a·tanθ·tanφ / (tanφ – tanθ).
Proof: Let height h, far point at distance d = h/tanθ, near point at distance (d-a), so h = (d-a)tanφ. Equating: d·tanθ = (d-a)tanφ => d(tanφ-tanθ) = a·tanφ => d = a·tanφ/(tanφ-tanθ). Then h = a·tanθ·tanφ / (tanφ – tanθ). Use this general template for any two-position elevation problem, substituting the problem’s own values of a, θ, φ. - Q7 (Multi-number word problem). A 1.5 m tall girl spots a balloon at 91.5 m height; as the balloon drifts away horizontally, the angle of elevation reduces from 60° to 30°. Find the distance the balloon travelled.
Height above eye = 90 m. d1 = 90/tan60° = 30√3. d2 = 90/tan30° = 90√3. Distance = 90√3-30√3 = 60√3 m (approx 103.9 m).
