NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry — Exercise Solutions

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Complete NCERT Solutions for CBSE Class 10 Maths Chapter 9 — Some Applications of Trigonometry: real-world heights and distances problems using angles of elevation and depression. Always draw the right triangle first.

Key Concepts (Quick Recap)

• Angle of elevation: measured upward from the horizontal.
• Angle of depression: measured downward from the horizontal.
• Choose the ratio that links the known and unknown sides — usually tan = height/base.
• Useful values: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3.

Exercise 9.1 — Heights and Distances

Q1. The angle of elevation of the top of a tower from a point 30 m from its base is 30°. Find the height of the tower. tan 30° = h/30 → h = 30 × (1/√3) = 30/√3 = 10√3 m ≈ 17.32 m.

Q2. A ladder leans against a wall making a 60° angle with the ground. The foot of the ladder is 2.5 m from the wall. Find the length of the ladder. cos 60° = 2.5/L → 1/2 = 2.5/L → L = 5 m.

Q3. A kite is flying at a height of 60 m. The string makes a 60° angle with the ground. Assuming no slack, find the length of the string. sin 60° = 60/L → (√3/2) = 60/L → L = 120/√3 = 40√3 m ≈ 69.28 m.

Q4. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower. Let the horizontal distance be d. Depression of foot: tan 45° = 7/d → d = 7 m. Elevation of top: tan 60° = h'/d → h' = 7√3 (the part above the building's top). Total tower height = 7 + 7√3 = 7(1 + √3) m ≈ 19.12 m.

Q5. The angle of elevation of the top of a tower from a point on the ground is 45°. On walking 20 m towards the tower, it becomes 60°. Find the height of the tower. Let height = h, far point distance = x. tan 45° = h/x → x = h. Near point: tan 60° = h/(x − 20) → √3 = h/(h − 20) → √3(h − 20) = h → h(√3 − 1) = 20√3 → h = 20√3/(√3 − 1). Rationalise: h = 20√3(√3 + 1)/[(√3)² − 1] = 20√3(√3 + 1)/2 = 10√3(√3 + 1) = 10(3 + √3) = 30 + 10√3 m ≈ 47.32 m.

Q6. An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from her eyes is 45°. Find the height of the tower. From eye level: tan 45° = h/28.5 → h = 28.5 m above the eyes. Total tower height = 28.5 + 1.5 = 30 m.

Q7. A tree breaks due to a storm and the broken part bends so that the top touches the ground, making a 30° angle with the ground. The distance from the foot of the tree to the point where the top touches the ground is 8 m. Find the original height of the tree. Let the broken (slant) part = L and the standing part = p. cos 30° = 8/L → L = 8/(√3/2) = 16/√3. tan 30° = p/8 → p = 8/√3. Original height = p + L = 8/√3 + 16/√3 = 24/√3 = 8√3 m ≈ 13.86 m.

Common Mistakes

• Confusing the angle of elevation with the angle of depression (both measured from the horizontal).
• Forgetting to add the observer's/building's height when the problem gives it.
• Not rationalising the denominator in the final answer.

Exam Tips

• Always draw a labelled right-triangle diagram before writing equations.
• The angle of depression from the top equals the angle of elevation from the bottom (alternate angles).
• Leave answers in simplest surd form unless a decimal is asked (use √3 ≈ 1.732).