NCERT Solutions for Class 10 Maths Chapter 9: Some Applications of Trigonometry (NCERT 2026–27)

These Class 10 Maths Chapter 9 solutions cover Some Applications of Trigonometry — the chapter on heights and distances. Every question of Exercise 9.1 is reproduced verbatim from the NCERT textbook (Reprint 2026–27) and solved step by step, using the angle of elevation and angle of depression with right-triangle trigonometric ratios. Each answer is cross-checked against the book’s answer key.

Class: 10 Subject: Mathematics Chapter: 9 Title: Some Applications of Trigonometry Exercise: 9.1 (15 questions) Session: 2026–27
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Chapter 9 Overview

Chapter 9, Some Applications of Trigonometry, shows how the trigonometric ratios learnt in Chapter 8 are used in real life to find the heights of objects (towers, poles, buildings, trees, balloons) and distances between points that cannot be measured directly. The key idea is to draw the situation as a right-angled triangle, identify the line of sight, the angle of elevation or angle of depression, and then pick the trigonometric ratio (sin, cos or tan) that connects the side you know to the side you want. Standard angles 30°, 45° and 60° appear throughout, so their exact ratio values are used in every solution below.

Key Concepts & Definitions

Line of sight: the straight line drawn from the eye of an observer to the point on the object being viewed.

Horizontal level: the horizontal line through the observer’s eye.

Angle of elevation: the angle formed by the line of sight with the horizontal when the object is above the horizontal level (we raise our head to look up).

Angle of depression: the angle formed by the line of sight with the horizontal when the object is below the horizontal level (we lower our head to look down).

Alternate-angle fact: an angle of depression from the top of an object equals the angle of elevation of that top from the lower point, because the horizontal lines are parallel and the line of sight is a transversal.

Important Formulas & Ratios (Chapter 9)

Right-triangle ratios: sinθ = opposite / hypotenuse  •  cosθ = adjacent / hypotenuse  •  tanθ = opposite / adjacent.

Standard values used here:

tan 30° = 1/√3,   tan 45° = 1,   tan 60° = √3.

sin 30° = 1/2,   sin 45° = 1/√2,   sin 60° = √3/2.

cos 30° = √3/2,   cos 45° = 1/√2,   cos 60° = 1/2.

Method: draw a right triangle → mark the known angle and side → use the ratio that links the known side to the required side → solve. (Take √3 ≈ 1.732 when a decimal is asked.)

Exercise 9.1 — Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book. All figures are described in words.

1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11).

SOLUTION Let AB be the vertical pole (height = h) and AC the rope of length 20 m, making 30° with the ground at C. Triangle ABC is right-angled at B (foot of pole). Here the height AB is opposite the 30° angle and the rope AC is the hypotenuse, so use sine. sin 30° = AB / AC ⇒ 1/2 = h / 20. ∴ h = 20 × 1/2 = 10 m. The height of the pole is 10 m.

2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

SOLUTION Let the tree break at B. The lower (standing) part AB is vertical; the broken part BC bends over and its top touches the ground at C, with the foot of the tree at A. So ∠ACB = 30° and AC = 8 m, with the right angle at A. The standing part AB is opposite 30° and AC is adjacent, so: tan 30° = AB / AC ⇒ 1/√3 = AB / 8 ⇒ AB = 8/√3 m. The broken part BC is the hypotenuse: cos 30° = AC / BC ⇒ √3/2 = 8 / BC ⇒ BC = 16/√3 m. Total height = AB + BC = 8/√3 + 16/√3 = 24/√3 = 8√3 m. ∴ height of the tree = 8√3 m (≈ 13.86 m).

3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

SOLUTION In each case the slide is the hypotenuse and the given height is opposite the angle, so use sine: length = height / sinθ. Younger children: sin 30° = 1.5 / L ⇒ 1/2 = 1.5 / L ⇒ L = 1.5 × 2 = 3 m. Elder children: sin 60° = 3 / L ⇒ √3/2 = 3 / L ⇒ L = 3 × 2/√3 = 6/√3 = 2√3 m (≈ 3.46 m).

4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

SOLUTION Let AB be the tower (height h) and C the point on the ground 30 m from the foot B, with ∠ACB = 30°, right-angled at B. tan 30° = AB / BC ⇒ 1/√3 = h / 30. ∴ h = 30/√3 = 30√3/3 = 10√3 m (≈ 17.32 m).

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

SOLUTION Let AB be the height of the kite (60 m) and AC the string of length L tied at point C, with ∠ACB = 60°, right-angled at B. The height is opposite 60° and the string is the hypotenuse: sin 60° = AB / AC ⇒ √3/2 = 60 / L. ∴ L = 60 × 2/√3 = 120/√3 = 40√3 m (≈ 69.28 m).

6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

SOLUTION The boy’s eyes are 1.5 m above the ground, so the height of the building above eye level is 30 − 1.5 = 28.5 m. Call this PQ, with the eye level at Q. Let the first position be at horizontal distance d₁ (angle 30°) and the second at d₂ (angle 60°). tan 30° = 28.5 / d₁ ⇒ 1/√3 = 28.5 / d₁ ⇒ d₁ = 28.5√3 m. tan 60° = 28.5 / d₂ ⇒ √3 = 28.5 / d₂ ⇒ d₂ = 28.5/√3 = 9.5√3 m. Distance walked = d₁ − d₂ = 28.5√3 − 9.5√3 = 19√3 m. ∴ the boy walked 19√3 m (≈ 32.9 m) towards the building.

7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

SOLUTION Let the building be AB = 20 m and the tower be BC on top of it. Let P be the point on the ground at distance AP = d from the foot A. The bottom of the tower B subtends 45° and the top C subtends 60°. From the bottom: tan 45° = AB / AP ⇒ 1 = 20 / d ⇒ d = 20 m. From the top: tan 60° = AC / AP ⇒ √3 = (20 + BC) / 20 ⇒ 20 + BC = 20√3. ∴ BC = 20√3 − 20 = 20(√3 − 1) m = 20(√3 − 1) m (≈ 14.64 m).

8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

SOLUTION Let the pedestal be AB = h and the statue be BC = 1.6 m on top of it. Let P be the point on the ground with AP = d. Top of pedestal B subtends 45°, top of statue C subtends 60°. From the top of the pedestal: tan 45° = h / d ⇒ 1 = h / d ⇒ d = h. From the top of the statue: tan 60° = (h + 1.6) / d ⇒ √3 = (h + 1.6) / h. √3 h = h + 1.6 ⇒ h(√3 − 1) = 1.6 ⇒ h = 1.6 / (√3 − 1). Rationalise: h = 1.6(√3 + 1) / [(√3 − 1)(√3 + 1)] = 1.6(√3 + 1) / 2 = 0.8(√3 + 1) m. ∴ height of the pedestal = 0.8(√3 + 1) m (≈ 2.19 m).

9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

SOLUTION Let the building be AB (height h) and the tower be CD = 50 m, standing distance d apart on level ground (B is the foot of the building, D is the foot of the tower). From the foot of the building, the top of the tower has elevation 60°: tan 60° = 50 / d ⇒ √3 = 50 / d ⇒ d = 50/√3 m. From the foot of the tower, the top of the building has elevation 30°: tan 30° = h / d ⇒ 1/√3 = h / (50/√3). h = (1/√3) × (50/√3) = 50/3 m. ∴ height of the building = 50/3 m (≈ 16.67 m).

10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

SOLUTION Let each pole have height h. Let the point P be at distance x from the foot of the first pole (angle 60°); then it is (80 − x) from the second pole (angle 30°). First pole: tan 60° = h / x ⇒ √3 = h / x ⇒ h = √3 x.   (1) Second pole: tan 30° = h / (80 − x) ⇒ 1/√3 = h / (80 − x) ⇒ h = (80 − x)/√3.   (2) Equate (1) and (2): √3 x = (80 − x)/√3 ⇒ 3x = 80 − x ⇒ 4x = 80 ⇒ x = 20 m. So the point is 20 m from the first pole and 80 − 20 = 60 m from the second pole. Height h = √3 × 20 = 20√3 m (≈ 34.64 m).

11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

SOLUTION Let AB be the tower of height h, with B at the foot on one bank. Let C be the point on the opposite bank directly across (angle 60°), so BC = w is the width of the canal. Let D be 20 m further back from C, so BD = w + 20 (angle 30°). From C: tan 60° = h / w ⇒ √3 = h / w ⇒ h = √3 w.   (1) From D: tan 30° = h / (w + 20) ⇒ 1/√3 = h / (w + 20) ⇒ h = (w + 20)/√3.   (2) Equate: √3 w = (w + 20)/√3 ⇒ 3w = w + 20 ⇒ 2w = 20 ⇒ w = 10 m. Width of the canal = 10 m; height h = √3 × 10 = 10√3 m (≈ 17.32 m).

12. 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°. Determine the height of the tower.

SOLUTION Let the building be AB = 7 m and the cable tower be CD, the two standing distance d apart. From the top A of the building, draw a horizontal line meeting the tower at E (so AE = d, BD = AE = d). E is at the same height (7 m) as A. Angle of depression of the foot D is 45° ⇒ the horizontal distance equals the drop: in right triangle ABD, tan 45° = AB / d ⇒ 1 = 7 / d ⇒ d = 7 m. So BD = AE = 7 m, and the foot of the tower D is level with B. Angle of elevation of the top C is 60°: in right triangle AEC, tan 60° = CE / AE ⇒ √3 = CE / 7 ⇒ CE = 7√3 m. Total height of tower CD = CE + ED = 7√3 + 7 (since ED = AB = 7 m). ∴ height of the tower = 7(√3 + 1) m (≈ 19.12 m).

13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

SOLUTION Let AB be the lighthouse, AB = 75 m, with B at sea level. The nearer ship (larger depression 45°) is at C and the farther ship (smaller depression 30°) is at D, with B, C, D in a line. Angles of depression equal the angles of elevation from the ships. Nearer ship: tan 45° = AB / BC ⇒ 1 = 75 / BC ⇒ BC = 75 m. Farther ship: tan 30° = AB / BD ⇒ 1/√3 = 75 / BD ⇒ BD = 75√3 m. Distance between ships = CD = BD − BC = 75√3 − 75 = 75(√3 − 1) m. ∴ the distance between the two ships = 75(√3 − 1) m (≈ 54.9 m).

14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

SOLUTION The girl’s eyes are 1.2 m above the ground, so the height of the balloon above eye level is 88.2 − 1.2 = 87 m (the balloon moves along a horizontal line at this height). Let the horizontal distance from the girl at the first instant (angle 60°) be x₁ and at the second instant (angle 30°) be x₂. tan 60° = 87 / x₁ ⇒ √3 = 87 / x₁ ⇒ x₁ = 87/√3 = 29√3 m. tan 30° = 87 / x₂ ⇒ 1/√3 = 87 / x₂ ⇒ x₂ = 87√3 m. Distance travelled = x₂ − x₁ = 87√3 − 29√3 = 58√3 m. ∴ the balloon travelled 58√3 m (≈ 100.46 m).

15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

SOLUTION Let the tower be AB of height h, with B at the foot. The car is first at C (depression 30°, so elevation 30° from the car) and 6 seconds later at D (depression 60°). tan 30° = h / BC ⇒ 1/√3 = h / BC ⇒ BC = h√3. tan 60° = h / BD ⇒ √3 = h / BD ⇒ BD = h/√3. Distance covered in 6 s = CD = BC − BD = h√3 − h/√3 = (3h − h)/√3 = 2h/√3. Remaining distance to foot = BD = h/√3. Since speed is uniform, time ∝ distance. Time for BD / Time for CD = (h/√3) / (2h/√3) = 1/2 ⇒ time for BD = (1/2) × 6 = 3 s. ∴ the car takes 3 seconds to reach the foot of the tower.

Common Mistakes to Avoid

Watch out for these

  • Choosing the wrong ratio — match the known and required sides: opposite + hypotenuse → sine; adjacent + hypotenuse → cosine; opposite + adjacent → tangent.
  • Forgetting the observer’s height — for the boy (Q6) and the girl (Q14), subtract eye height (1.5 m, 1.2 m) from the object’s height first.
  • Confusing angle of elevation with angle of depression — remember the depression from a top equals the elevation from the lower point (alternate angles).
  • Leaving an irrational denominator — rationalise, e.g. 30/√3 = 10√3, and 1.6/(√3 − 1) = 0.8(√3 + 1).
  • Not labelling a clean diagram — an unmarked figure leads to wrong opposite/adjacent sides.
  • Mixing up the near and far objects — the larger angle is always for the nearer point.

Practice MCQs & Assertion–Reason

1. The angle formed by the line of sight with the horizontal when the object is below the horizontal level is called the angle of:

(a) elevation    (b) depression    (c) reflection    (d) projection

2. A 20 m long rope tied from the top of a pole to the ground makes 30° with the ground. The height of the pole is:

(a) 10 m    (b) 10√3 m    (c) 20√3 m    (d) 40 m

3. The angle of elevation of the top of a tower from a point 30 m from its foot is 30°. The height of the tower is:

(a) 30√3 m    (b) 10√3 m    (c) 15 m    (d) 60 m

4. A kite flies at a height of 60 m; its string makes 60° with the ground. The length of the string is:

(a) 30√3 m    (b) 40√3 m    (c) 60√3 m    (d) 120 m

5. An angle of depression of a point equals the angle of elevation of the observer’s position from that point because the two horizontal lines are:

(a) perpendicular    (b) parallel    (c) equal in length    (d) concurrent

6. The shadow of a tower is shortest when the Sun’s altitude (angle of elevation) is:

(a) 30°    (b) 45°    (c) 60°    (d) 0°

7. From a point on the ground the angle of elevation of the top of a 20 m building is 45°. The distance of the point from the foot is:

(a) 10 m    (b) 20 m    (c) 20√3 m    (d) 40 m

8. As observed from a 75 m lighthouse, two ships in line have depressions 30° and 45°. The distance between them is:

(a) 75 m    (b) 75√3 m    (c) 75(√3 − 1) m    (d) 150 m

9. A tower casts a shadow such that the angle of elevation of the Sun is 45°. Then the height of the tower equals its:

(a) half the shadow    (b) shadow length    (c) twice the shadow    (d) √3 × shadow

10. To find the length of a slide that reaches a height of 3 m at 60°, the correct ratio to use is:

(a) cos 60°    (b) tan 60°    (c) sin 60°    (d) cot 60°

Answer key: 1-(b), 2-(a), 3-(b), 4-(b), 5-(b), 6-(c), 7-(b), 8-(c), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: When the observed point is above the horizontal, the angle of the line of sight with the horizontal is the angle of elevation.

Reason: The line of sight is drawn from the eye of the observer to the point being viewed.

A-R 2. Assertion: The angle of depression of a point from a tower top equals the angle of elevation of the tower top from that point.

Reason: The horizontal at the top and the horizontal at the point are parallel, so the line of sight makes equal alternate angles.

A-R 3. Assertion: A 20 m rope at 30° to the ground reaches a pole top 10 m high.

Reason: sin 30° = 1/2, and height = rope × sin 30°.

A-R 4. Assertion: The height of a tower equals the length of its shadow when the Sun’s altitude is 30°.

Reason: tan 30° = 1, so height equals shadow at 30°.

A-R 5. Assertion: For a car approaching a tower at uniform speed, equal time intervals cover equal horizontal distances.

Reason: For uniform motion, distance is directly proportional to time.

Answer key: 1-(A), 2-(B), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • The line of sight runs from the observer’s eye to the object; with the horizontal it makes the angle of elevation (object above) or depression (object below).
  • Heights and distances are found by modelling the situation as a right triangle and choosing sin, cos or tan to link the known side to the required side.
  • An angle of depression equals the corresponding angle of elevation (alternate angles between parallel horizontals).
  • Use exact values: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3; rationalise irrational denominators.
  • Subtract the observer’s eye height from an object’s height before using the angle (boy 1.5 m, girl 1.2 m).
  • For two-angle problems, set up one equation per angle and eliminate the unknown distance.
  • For uniform-speed problems, time is proportional to distance along the line of motion.

How to score full marks in this chapter

Always start with a clean, labelled diagram showing the right angle, the given angle, and the known side — many marks are awarded for the correct figure and ratio choice. State the trigonometric ratio with its formula before substituting, keep √3 in surd form until the last step, then approximate only if a decimal is required (√3 ≈ 1.732). For problems with two angles, write both equations clearly and label them (1) and (2) before eliminating the unknown distance.

Frequently Asked Questions

What is Class 10 Maths Chapter 9 about?

Chapter 9, Some Applications of Trigonometry, applies trigonometric ratios to real-life problems of heights and distances — finding the height of towers, poles, buildings and balloons, and distances between objects — using the angle of elevation and angle of depression in right triangles.

How many exercises are there in Class 10 Maths Chapter 9?

There is one exercise, Exercise 9.1, containing 15 questions. All 15 are solved step by step on this page, with answers verified against the NCERT answer key.

What is the difference between angle of elevation and angle of depression?

The angle of elevation is the angle the line of sight makes with the horizontal when you look up at an object above eye level; the angle of depression is the angle when you look down at an object below eye level. They are equal alternate angles when the two horizontals are parallel.

Are these Class 10 Maths Chapter 9 solutions free?

Yes. All solutions are free and follow the official NCERT textbook for the 2026–27 session, with every answer checked against the book’s answer key.

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