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CBSE Maths · Class 10

Chapter 9: Some Applications of Trigonometry

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Date: ______________
Marks: ____ / 10

Real-life problems on heights and distances using angles of elevation and depression.

Questions

  1. Ladder 1010 m at 6060^\circ: height reached.
  2. 2020 m tower, elevation 4545^\circ: distance from base.
  3. Cliff 100100 m, boat at 3030^\circ depression.
  4. Pole 1515 m, sun 4545^\circ: shadow length.
  5. Tree 2020 m tall, sun at 6060^\circ: shadow length.
  6. Tower shadow 3030 m, sun 3030^\circ: height.
  7. Angles of elevation from a point 1010 m and 4040 m away are complementary. Find height.
  8. From top of a 5050 m tower, angle of depression of two objects on ground on same side are 3030^\circ and 6060^\circ. Find distance between them.
  9. A kite is flying at height 6060 m; string makes 6060^\circ with ground. Length of string?
  10. A ladder 3\sqrt 3 m leans against a wall; foot is 11 m from the wall. Angle with ground?

Extra Practice (Optional)

  • A man at 3030 m sees the top of a tower at 6060^\circ. Find tower's height.
  • From a boat, an observer sees lighthouse top at 4545^\circ; sailing closer 5050 m, it becomes 6060^\circ. Height?
  • Two towers of equal height, 6060 m apart. From middle, elevations 3030^\circ and 6060^\circ. Heights?
  • Balloon at height 8888 m; elevation from a point on ground 3030^\circ; find distance.
  • Aeroplane 30003000 m high; elevation changes from 6060^\circ to 3030^\circ in 3030 s. Speed?

Answer Key (For Teachers)

  1. Answer: 535\sqrt 3 m.
    Working: 10sin6010\sin 60.
  2. Answer: 2020 m.
    Working: tan45=1\tan 45 = 1.
  3. Answer: 1003100\sqrt 3 m.
    Working: tan30=100/d\tan 30 = 100/d.
  4. Answer: 1515 m.
    Working: tan45\tan 45: shadow == height.
  5. Answer: 20/320/\sqrt 3 m.
    Working: x=20/3x = 20/\sqrt 3.
  6. Answer: 10310\sqrt 3 m.
    Working: h=30/3h = 30/\sqrt 3.
  7. Answer: 2020 m.
    Working: Let angle from 1010 m be θ\theta; from 4040 m be 90θ90-\theta.h=10tanθ=40cotθh = 10\tan\theta = 40\cot\theta, so tanθ=2\tan\theta = 2; h=102=20h = 10\cdot 2 = 20.
  8. Answer: 100/3100/\sqrt 3 m.
    Working: Distances from base: 50/tan60=50/350/\tan 60 = 50/\sqrt 3 and 50/tan30=50350/\tan 30 = 50\sqrt 3.Difference: 50350/3=50(31/3)=100/350\sqrt 3 - 50/\sqrt 3 = 50(\sqrt 3 - 1/\sqrt 3) = 100/\sqrt 3.
  9. Answer: 40340\sqrt 3 m.
    Working: sin60=60/LL=602/3=403\sin 60 = 60/L \Rightarrow L = 60\cdot 2/\sqrt 3 = 40\sqrt 3.
  10. Answer: 54.7\approx 54.7^\circ.
    Working: cosθ=1/3\cos\theta = 1/\sqrt 3; or sinθ=h/3\sin\theta = h/\sqrt 3 with h=31=2h = \sqrt{3-1} = \sqrt 2; tanθ=2\tan\theta = \sqrt 2; θ54.7\theta \approx 54.7^\circ.
Gyaan Maths · CBSE Class 10 · Chapter 9
Gyaan Maths — CBSE Class 6-10, built for board-toppers.
v1.1 · Feb 2026