---
source: ncert_exercises
topic: trigonometry
class_level: class_10
chapter: ch9
exercise: ex9.1
difficulty: intermediate
---

Class 10 | Ch9: Some Applications of Trigonometry | Exercise 9.1

Key Terms:
Angle of Elevation: angle above horizontal to look UP at object.
Angle of Depression: angle below horizontal to look DOWN at object.

Basic Method:
1. Draw diagram. Label known and unknown.
2. Identify right triangle.
3. Apply tan, sin or cos as needed.
4. Solve for unknown.

Common tan values: tan30°=1/√3, tan45°=1, tan60°=√3.

Q1. A circus artist climbs a rope 20m long. If rope makes 30° with ground, find height of pole.
Answer: sin30°=h/20 → h=20×1/2=10m. ✅

Q2. A tree breaks and top touches ground 8m from base at angle 60°. Find height of tree.
Answer: tan60°=vertical/8 → vertical=8√3m. Broken part=8/cos60°=16m. Total height=8√3+16m. ✅

Q3. Contractor wants to erect a pole. He observes top at angle 60° from 30m away. Find height.
Answer: tan60°=h/30 → h=30√3m. ✅

Q4. The angle of elevation of top of tower from point 30m away is 30°. Find height.
Answer: tan30°=h/30 → h=30×(1/√3)=10√3m. ✅

Q5. A kite is flying at height h. String makes 60° with ground. If string=60m, find h.
Answer: sin60°=h/60 → h=60×√3/2=30√3m. ✅

Q6. 1.5m tall boy stands 28.5m from chimney of 30m height. Find angle of elevation.
Answer: Effective height=30-1.5=28.5m. tanθ=28.5/28.5=1 → θ=45°. ✅

Q7. From a point P, angle of elevation of balloon=60°. After P walks 100m, angle=30°. Find height of balloon.
Answer: Let height=h, first distance=d. tan60°=h/d → h=d√3. tan30°=h/(d+100) → h=(d+100)/√3. d√3=(d+100)/√3 → 3d=d+100 → d=50m. h=50√3m. ✅

Q8. Shadow of 6m tall pole is 2√3m long. Find angle of elevation of sun.
Answer: tanθ=6/(2√3)=3/√3=√3 → θ=60°. ✅

Q9. Angle of depression of two ships from top of lighthouse (75m) are 30° and 45°. Find distance between ships (on same side).
Answer: d₁=75/tan30°=75√3m. d₂=75/tan45°=75m. Distance=75√3-75=75(√3-1)m. ✅

Q10. From top of 50m tower, angles of depression of top and bottom of building are 30° and 45°. Find height of building.
Answer: d=50/tan45°=50m (from bottom). Height building=50-50tan30°=50-50/√3=50(1-1/√3)=50(√3-1)/√3m. ✅

Q11. A TV tower stands on top of building. Tower subtends angle 'a' at point on ground, building subtends angle 'b'. Find height of tower.
Answer: Let building height=h, tower=t, distance from point=d. tanb=h/d, tan(a+b)=(h+t)/d. t=d(tan(a+b)-tanb). ✅

Q12. From top of hill, angles of depression of two cars A and B are 60° and 30°. If A is between B and hill, find distance AB in terms of height h.
Answer: Let hill height=h. Distance to A: h/tan60°=h/√3. Distance to B: h/tan30°=h√3. AB=h√3-h/√3=h(3-1)/√3=2h/√3=2h√3/3. ✅

Q13. Angles of elevation of top of vertical tower from points P and Q at distances a and b respectively (on same horizontal line) are complementary. Prove height=√(ab).
Answer: Let θ and (90°-θ) be angles. h=a tanθ and h=b tan(90°-θ)=b cotθ. Multiply: h²=ab → h=√(ab). ✅

Q14. Vertical pole 6m high and tower some distance apart. Angle of elevation from bottom of pole=60°, from top=45°. Find distance and height of tower.
Answer: Let distance=d, height=H. tan60°=H/d → H=d√3. tan45°=(H-6)/d → H-6=d. d√3-6=d → d(√3-1)=6 → d=6/(√3-1)=3(√3+1). H=3√3(√3+1)=3(3+√3)=9+3√3m. ✅

Q15. From the top of a 7m high building, angle of elevation of cable tower top=60° and angle of depression of its foot=45°. Find height of tower.
Answer: Let distance=d. tan45°=7/d → d=7m. tan60°=h'/7 where h' is extra height above building. h'=7√3. Total height=7+7√3=7(1+√3)m. ✅

Q16. The angles of elevation of a jet from two points P and Q are 60° and 30°. If jet is directly above midpoint of PQ and PQ=2000m, find height.
Answer: Midpoint=1000m from each. tan60°=h/1000 → h=1000√3m (from P). Verify from Q: tan30°=h/3000=1000√3/3000=1/√3 ✓. Height=1000√3m≈1732m. ✅