Some Applications of Trigonometry is Chapter 9 of Class 10 NCERT Maths (Reprint 2026-27). It uses three core ideas: line of sight, angle of elevation, and angle of depression to find heights and distances without direct measurement. Every problem reduces to identifying a right triangle, picking the correct trigonometric ratio, and solving for the unknown side.
Important questions class 10 maths chapter 9 on this page are mapped to the NCERT textbook Reprint 2026-27, arranged by question type, and include full solutions so you can check your working at every step.
A student who draws the diagram correctly before writing a single equation scores full marks on every question in this chapter. Most board marks are lost on diagram errors. A student who uses tan when the hypotenuse is given, or forgets the observer's height, loses marks that are otherwise easy to get. All questions and solutions are available section by section below.
Key Takeaways
| Topic |
What to Know |
| Line of Sight |
Imaginary line from observer's eye to the object |
| Angle of Elevation |
Angle above horizontal when looking up at an object |
| Angle of Depression |
Angle below horizontal when looking down at an object |
| Primary Ratio Used |
tan theta = Height / Distance (most common in this chapter) |
| Standard Angles |
30°, 45°, 60°: memorise sin, cos, tan values |
| Observer Height |
Subtract from total height before setting up the triangle |
| Alternate Angles |
Angle of depression = angle of elevation at base (alternate interior angles) |
| Chapter Scope |
Heights and Distances only: no identity proofs |
Introduction to Some Applications of Trigonometry
Trigonometry ratios from Chapter 8 are tools. Chapter 9 puts those tools to work on real-world problems involving heights and distances.
Three terms run through every question in this chapter. Line of sight is the straight line from an observer's eye to the point being viewed. Angle of elevation is the angle the line of sight makes with the horizontal when the object is above the observer's level. Angle of depression is the angle the line of sight makes with the horizontal when the object is below the observer's level.
One important geometry rule: the angle of depression from a height equals the angle of elevation from the base. This comes from alternate interior angles with the horizontal. Mark this correctly in your diagram or you will get the wrong triangle.
Important Questions Class 10 Maths All Chapters
Important Questions on Heights and Distances
Heights and Distances is the only topic in Chapter 9. Every question from the simplest 2-mark problem to the longest 5-mark board question is a heights-and-distances problem. These height and distance class 10 important questions are organised by the number of triangles involved and what information is given or hidden.
The standard values you must know:
| Angle |
sin |
cos |
tan |
| 30° |
1/2 |
root3/2 |
1/root3 |
| 45° |
1/root2 |
1/root2 |
1 |
| 60° |
root3/2 |
1/2 |
root3 |
Always use tan theta when height and distance are the two sides in the problem. Switch to sin or cos only when the hypotenuse (rope, ladder, string) is given or required.
Important Questions on Line of Sight, Angle of Elevation and Angle of Depression
These are the definitional and direct-application questions. They carry 1 to 2 marks each in board exams. These class 10 maths chapter 9 important questions test whether students can set up the triangle correctly from a word problem.
Q1. A tower stands vertically on the ground. From a point 15 m away from the foot, the angle of elevation of the top is 60°. Find the height of the tower. Let AB = height of tower, BC = 15 m, angle ACB = 60°. tan 60° = AB/BC → root3 = AB/15 → AB = 15root3 m.
Q2. From a point on the ground, the angle of elevation of a chimney's top is 45°. The observer is 1.5 m tall and stands 28.5 m from the chimney. Find the total height. Let AE = height above observer's eye level, DE = 28.5 m. tan 45° = 1: AE = 28.5 m. Total height = 28.5 + 1.5 = 30 m.
Very Short Answer Questions on Applications of Trigonometry
One-mark and two-mark questions test direct ratio application. These applications of trigonometry class 10 extra questions are common in unit tests and CBSE 2026 board MCQ sets.
Q3. A kite flies at 60 m above the ground. The string makes 60° with the ground. Find the string's length. sin 60° = 60/AC → root3/2 = 60/AC → AC = 120/root3 = 40root3 m.
Q4. The angle of elevation of the top of a tower from 30 m away is 30°. Find the tower's height. tan 30° = h/30 → 1/root3 = h/30 → h = 10root3 m.
Q5. A circus artist climbs a 20 m rope tied from a pole top to the ground. The rope makes 30° with the ground. Find the pole's height. sin 30° = AB/AC → 1/2 = AB/20 → AB = 10 m.
Short Answer Questions on Applications of Trigonometry
Two-mark and three-mark questions test formula application with one or two setup steps. These some applications of trigonometry class 10 important questions cover the most tested short-answer formats in CBSE 2026 papers.
Q6. An electrician needs to reach a point 1.3 m below the top of a 5 m pole. She places a ladder at 60° to the horizontal. Find the ladder's length. (Use root3 = 1.73) BD = 5 - 1.3 = 3.7 m. sin 60° = BD/BC → root3/2 = 3.7/BC → BC = 3.7 x 2/root3 = 4.28 m (approx). Distance from pole foot = DC = 3.7/root3 = 2.14 m (approx).
Q7. A 1.5 m tall boy stands some distance from a 30 m tall building. His angle of elevation increases from 30° to 60° as he walks toward it. Find the distance walked. Effective height = 30 - 1.5 = 28.5 m. At 30°: BD = 28.5root3 m. At 60°: BC = 28.5/root3 = 28.5root3/3 m. Distance walked = BD - BC = 28.5root3 x 2/3 = 19root3 m.
Q8. The angle of elevation of the top of a building from a tower's foot is 30°. The angle of elevation of the tower's top from the building's foot is 60°. The tower is 50 m high. Find the building's height. BC = 50/root3 (from the tower triangle). AB = BC/root3 = 50/3 m.
Long Answer Questions on Applications of Trigonometry
Four-mark and five-mark questions each involve two right triangles and a two-step solution. These class 10 applications of trigonometry important questions carry the highest marks in CBSE 2026 board papers.
Q9. From the ground, angles of elevation of the bottom and top of a transmission tower fixed on a 20 m building are 45° and 60° respectively. Find the tower's height. In triangle BCD: tan 45° = 20/CD → CD = 20 m. In triangle ACD: tan 60° = AC/20 → AC = 20root3. Tower height = AC - BC = 20root3 - 20 = 20(root3 - 1) m.
Q10. A 1.6 m statue stands on a pedestal. From a point on the ground, the angle of elevation to the statue top is 60° and to the pedestal top is 45°. Find the pedestal height. In triangle BCD: tan 45° = BC/CD → BC = CD. In triangle ACD: tan 60° = AC/CD → root3 x BC = 1.6 + BC. BC(root3 - 1) = 1.6 → BC = 1.6/(root3 - 1) = 1.6(root3 + 1)/2 = 0.8(root3 + 1) m.
Q11. From the top of a 7 m building, the angle of elevation to a cable tower's top is 60° and angle of depression to its foot is 45°. Find the tower's height. tan 45° = 7/BC → BC = 7 m = AD. tan 60° = DE/AD → DE = 7root3. Tower height = DE + CD = 7root3 + 7 = 7(root3 + 1) m.
Tower and Building Important Questions
These questions involve two observation points for the same tower or two towers observed from a common point. These important questions of height and distance class 10 format appear in almost every CBSE 2026 board paper.
Q12. The angle of elevation of the top of a tower from two points at 4 m and 9 m from its base (same line) are complementary. Prove the tower height is 6 m. Let angles be x and (90° - x). tan x = h/4 → h = 4 tan x ... (i). tan(90° - x) = h/9 → cot x = h/9 → h = 9 cot x ... (ii). Multiplying (i) and (ii): h2 = 36 → h = 6 m. Proved.
Q13. Two poles of equal height stand on either side of an 80 m wide road. From a point between them, angles of elevation to the pole tops are 60° and 30°. Find the pole height and the point's distance from each pole. Let OD = x, OB = 80 - x, height = h. tan 30° = h/x → h = x/root3 ... (i). tan 60° = h/(80 - x) → h = root3(80 - x) ... (ii). From (i) and (ii): x/root3 = root3(80 - x) → x = 3(80 - x) → x = 60. h = 60/root3 = 20root3 m. Distances: 60 m and 20 m.
Ladder, Rope and Pole Important Questions
In these questions the hypotenuse is given: the rope, wire, or ladder length. Use sin or cos, not tan. These applications of trigonometry class 10 important questions test the most common ratio-selection error in board exams.
Q14. A contractor installs two slides. For children under 5: top at 1.5 m height, inclined at 30°. For older children: top at 3 m, inclined at 60°. Find each slide's length. Slide 1: sin 30° = 1.5/AC → AC = 3 m. Slide 2: sin 60° = 3/PR → PR = 2root3 m.
Broken Tree and Bent Pole Questions
A tree breaks in a storm. The broken part bends and touches the ground, forming an angle. The total tree height = upright part + broken part. These height and distance class 10 extra questions appear regularly as 3-mark and 4-mark problems.
Q15. A tree breaks in a storm. The broken part makes 30° with the ground. The distance from the tree's foot to where the top touches is 8 m. Find the tree's original height. cos 30° = 8/AC → AC = 16/root3 ... (i). tan 30° = AB/8 → AB = 8/root3 ... (ii). Total height = AB + AC = 8/root3 + 16/root3 = 24/root3 = 8root3 m.
Class 10 Maths Height and Distance Important Questions: Canal, Lighthouse and River
These questions involve two horizontal distances from a single vertical height. These some applications of trigonometry class 10 extra questions with solutions are among the most searched board-level problems.
Q16. A TV tower stands on a canal bank. From the opposite bank, the angle of elevation is 60°. From a point 20 m further away, it is 30°. Find the tower height and canal width. tan 60° = AB/BC → AB = root3 BC ... (i). tan 30° = AB/(BC + 20) → AB = (BC + 20)/root3 ... (ii). From (i) and (ii): 3BC = BC + 20 → BC = 10 m. AB = 10root3 m.
Q17. From a bridge 3 m above river banks, the angles of depression of the two banks are 30° and 45°. Find the river's width. tan 30° = 3/AD → AD = 3root3 m. tan 45° = 3/BD → BD = 3 m. Width = AD + BD = 3root3 + 3 = 3(root3 + 1) m.
Shadow-Based Questions from Applications of Trigonometry
Shadow problems give two sun altitudes and one shadow length difference. These some applications of trigonometry important questions test two-equation simultaneous solution using tan.
Q18. A tower's shadow is 40 m longer when the sun's altitude is 30° than when it is 60°. Find the tower's height. Let height = h, shorter shadow = x. tan 60° = h/x → h = xroot3 ... (i). tan 30° = h/(x + 40) → h = (x + 40)/root3 ... (ii). From (i) and (ii): 3x = x + 40 → x = 20. h = 20root3 m.
Two Angles from One Point Questions
Q19. From the ground, the angle of elevation of the top of a 10 m building is 30°. A flagstaff sits on top and the angle of elevation to the flagstaff top is 45°. Find the flagstaff length and building distance. (Use root3 = 1.732) tan 30° = 10/AP → AP = 10root3 m (distance = 17.32 m). Let flagstaff = x. tan 45° = (10 + x)/10root3 → 10root3 = 10 + x → x = 10(root3 - 1) = 7.32 m.
Angle of Depression Questions on Ships, Lighthouse and River
Q20. From a 75 m lighthouse, angles of depression of two ships on the same side are 45° and 30°. Find the distance between the ships. tan 45° = 75/BC → BC = 75 m. tan 30° = 75/BD → BD = 75root3 m. Distance = BD - BC = 75(root3 - 1) m.
Q21. From the top of a multi-storeyed building, angles of depression of the top and bottom of an 8 m building are 30° and 45°. Find the height of the multi-storeyed building and distance between them. tan 45° = 1: PC = AC. tan 30°: PD/BD = 1/root3 → BD = PDroot3. Since AC = BD and DC = 8: PD + 8 = PDroot3 → PD = 8/(root3 - 1) = 4(root3 + 1) m. Height = PD + 8 = 4(3 + root3) m. Distance = 4(3 + root3) m.
Moving Observer and Changing Angle Questions
Q22. A man at the top of a tower observes a car at 30° depression. Six seconds later, the angle is 60°. Find the time for the car to reach the tower from that second point. tan 60° = AB/BC → BC = AB/root3. tan 30° = AB/BD → BD = ABroot3. CD = BD - BC = AB root3 - AB/root3 = 2AB/root3. BC = AB/root3 = CD/2. The car covers CD in 6 seconds. BC = CD/2, so time = 3 seconds.
Q23. A 1.2 m tall girl spots a balloon at 88.2 m moving horizontally. Angle of elevation is 60°, then reduces to 30°. Find the distance the balloon travelled. Effective height = 88.2 - 1.2 = 87 m. At 60°: CD = 87/root3 = 29root3 m. At 30°: CE = 87root3 m. Distance = CE - CD = 87root3 - 29root3 = 58root3 m.
Class 10 Maths Chapter 9 Extra Questions with Solutions
These class 10 maths chapter 9 extra questions go beyond the standard NCERT exercises and appear in pre-board and annual examinations for CBSE 2026.
Q24. From the top of a 60 m building, angles of depression of the top and bottom of a tower are 45° and 60°. Find the tower's height. (root3 = 1.73) tan 60° = 60/BC → BC = 60/root3 = 20root3. tan 45° = AB/BC → AB = 20root3. Tower height = 60 - AB = 60 - 20root3 = 60 - 34.6 = 25.4 m.
Q25. A man on a ship 10 m above water observes the top of a hill at 60° elevation and its base at 30° depression. Find the hill's distance and height. tan 30° = 10/BC → BC = 10root3 m (distance from ship = 17.3 m). tan 60° = DE/AD = DE/10root3 → DE = 30 m. Total hill height = 10 + 30 = 40 m.
How to Solve Class 10 Maths Chapter 9 Questions
Follow these steps on every application of trigonometry extra questions problem. These are the six steps that separate students who lose marks on diagram errors from those who score full marks.
Step 1: Draw the figure first. Show all heights as vertical lines and all distances as horizontal lines. Label every point. Step 2: Identify the right triangle. Mark the right angle clearly. Step 3: Mark the angle correctly. Angle of elevation is at the observer's ground position looking up. For depression, use alternate interior angles to bring it down to the base triangle. Step 4: Choose the ratio. If height and distance are the two sides, use tan. If the hypotenuse is involved, use sin or cos. Step 5: Use standard values. tan 30° = 1/root3, tan 45° = 1, tan 60° = root3. Write these on rough paper before you start. Step 6: Add or subtract observer height last. The effective triangle height = object height minus observer height. Add it back only at the final answer.