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Class 10 Maths Notes — Chapter 9: Some Applications of Trigonometry

Heights and distances — using trigonometry to solve real-world problems.

Detailed NCERT notes

  • Line of sight: line from observer's eye to the object being viewed.
  • Angle of elevation: angle above horizontal (looking up).
  • Angle of depression: angle below horizontal (looking down).
  • Horizontal at observer's eye is the reference for both.
  • In an angle of depression problem, the alternate-interior-angle at the object equals the angle of depression at the observer (parallel horizontals).
  • Solving procedure: (1) Draw a clear diagram with the horizontal, vertical, and line of sight. (2) Mark all given lengths and angles. (3) Identify the right triangle. (4) Choose tan for opposite/adjacent, sin for opposite/hypotenuse, cos for adjacent/hypotenuse. (5) Solve.
  • Two-triangle setups: object seen from two positions, or two objects seen from one position → set up two equations, then subtract or divide.
  • If observer has non-negligible height, adjust the vertical distance accordingly.
  • Bearing / direction problems: rare in Board, but interpret 'from the top of a tower' vs 'from the ground'.

Formulas & key results

  • tan(angle) = opposite / adjacent (most common)
  • sin(angle) = opposite / hypotenuse; cos = adjacent / hypotenuse

Mind map

  • Diagram → angle → right triangle → ratio choice
  • Elevation (up) vs Depression (down)
  • Two-triangle setups

Tricks & shortcuts

  • Always sketch first — mark opposite/adjacent clearly.
  • For depression problems, transfer the angle to the base of the right triangle using parallel horizontals.

Common mistakes to avoid

  • Confusing elevation with depression.
  • Ignoring height of the observer when it's given.
  • Using sin when the hypotenuse isn't involved.

Competency-based questions & answers

  1. Q. A tower's shadow is √3 times its height. Find the angle of elevation of the Sun.
    A. tan θ = h/(√3 h) = 1/√3 ⇒ θ = 30°.
  2. Q. From the top of a 100 m tower, angles of depression of two cars on the same side are 45° and 30°. Find the distance between the cars.
    A. d₁ = 100 (from 45°); d₂ = 100√3 (from 30°) ⇒ distance = 100(√3 − 1) ≈ 73.2 m.
  3. Q. A ladder leans against a wall making 60° with the ground. Foot is 2.5 m from wall. Find the length of the ladder.
    A. cos 60° = 2.5/L ⇒ L = 5 m.