CBSE Important Questions Class 10 Maths Ch9 Applications of Trigonometry 2025-26

The chapter Some Applications of Trigonometry is one of the most practical and engaging topics in Mathematics Syllabus Class 10. It shows how trigonometry, introduced in the previous chapter, is used to measure heights and distances that are otherwise difficult to find directly.

This chapter strengthens your ability to visualise real-world problems mathematically. It focuses on applying trigonometric ratios to right triangles formed by lines of sight, angles of elevation, and angles of depression.

The Applications of Trigonometry Important Questions for Class 10 Mathematics help you:

• Understand real-life uses of trigonometry.
• Practice height and distance problems.
• Strengthen the ability to draw and interpret mathematical diagrams accurately.

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IMPORTANT QUESTIONS CLASS 10

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(Most Important Questions of this Chapter from our 📕)

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Ch9 Some Applications of Trigonometry Important Questions class 10

1. Shown below is a rectangular tub of water of depth 34 cm. An object O is at the bottom of the tub. The image of the object is formed at I for an observer at Q.

(Note: The figure is not to scale.)

Find the distance by which the object seem to be moved for the observer. Show your work and give valid reasons.

(Note: Take √2 = 1.4, √3 = 1.7)

Soln. Real depth of the object: d=34 cm

Angle of incidence (i) = 45°

Angle of refraction (r) = 30°

Snell's Law: n₁sin⁡i=n₂sin⁡r

Here, n₁=1(air refractive index) and n₂=μ (water refractive index).

sin45°=μsin30°

Substituting values

0.7=μ×0.5

0.7/0.5=μ

1.4 =μ

apparent depth d_a= Real depth​/μ

apparent depth d_a= 34/1.4

≈24.3cm

The object appears to move by the difference between the real depth and apparent depth:

Shift=d−d_a​

Shift=34−24.3

=9.7cm.

The object seems to move upward by approximately 9.7 cm for the observer.

2. Answer the questions based on the information given below.

At an archery academy, Guru Drona had floated a gift box with two balloons at a height of H metres from the table. As part of his practice, Arjuna was given the task to bring the gift box to the table placed below. Arjuna was standing on the ground at a horizontal distance of 100 metres from the table at point B. He aimed at the balloons with an elevation angle of 9 and shot the arrow to burst one of the balloons.When Arjuna burst the first balloon, the box came down to the height of h metres from the table. He now reduced his angle of elevation by ẞ and shot his arrow at the second balloon. The second balloon burst and the gift box landed safely on the table. Assume that Arjuna's arrows travelled in straight lines and did not curve down.

(Use √3 = 1.73, √2 = 1.41)

2.1 If θ = 45° and ẞ = 15°, what is the difference between the box's initial height and its height after the first shot?

a. 100- 100 √3m

b. 100 √3m

c. 100√3 - 100 m

d. (cannot be calculated without knowing H.)

Soln. (c) 100√3 - 100 m

Explanation:

height H of the first balloon: 

H=100tan45°

=100.

height h after the first shot:

h=100tan15°

h=100(2−√3​)

Δh =H−h

=100−100(2−√3​)

Δh=100√3​​−100

2.2 If θ = 45° and ẞ = 15°, what is the distance that the arrow has to travel to burst the second balloon?

a. 100√3/2 m

b. 200/√3 m

c. 100√2 m

d. 100 √3m

Soln. (d) 100 √3m

Explanation: For the second balloon, at a height h:

h=100tan15°

100(2−√3)

slant distance d: h/sin15°

sin15°= (√6 - √2)/4

d= (100(2−√3))/((√6 - √2)/4)

d= 100 √3m

2.3 For Ashwatthama, Guru Drona raised the gift box further higher such that the angles θ and ẞ were 60° and 30° respectively. What is the value of the ratio H/h now?

a. 1/√3

b. √3 

c. 2

d. 3

Soln. (d) 3

Explanation: H=100tan60°

=100 √3

h=100tan30°

=100/√3​

H/h = 100 √3/100/√3

= 3

2.4 When the initial angle of elevation, θ, was 45°, Arjuna felt uncomfortable as it strained his neck. From his original spot, approximately how much should he retreat away from the balloons, so that the new angle of elevation, θ, becomes 30°?

a. 73 m

b. 100 m

c. 173 m

d. (cannot be calculated without knowing H.)

Soln. (a) 73m

Explanation: H=100tan45°

=100

the horizontal distance x is = H/tan 30°

x= 100/1√3

x= 100√3

Retreat=x−100

100√3 - 100

√3 ​≈1.73

Retreat=173−100

=73m.

2.5 Arjuna measured that θ = 45°. Right before he could shoot the first arrow, a gust of wind pushed the balloons 15 m higher. What should Arjuna do to ensure that he doesn't miss?

a. Move towards the table by 15 m but keep the arrow at the same angle of elevation.

b. Move away from the table by 15 m but keep the arrow at the same angle of elevation.

c. Increase the arrow's angle of elevation by 15° but stay at the same place.

d. Move away from the table by 15 m and increase the arrow's angle of elevation by 15°.

Soln. (a) Move towards the table by 15 m but keep the arrow at the same angle of elevation.

Explanation: With a rise of 15 m, the new height becomes H=115

To maintain the same angle of elevation θ=45° the horizontal distance must equal the new height (x=115)

Arjuna must move towards the table by.

Δx=115−100

=15m

Move towards the table by 15 m but keep the arrow at the same angle of elevation.

3. The position of an eagle and two identical geese are shown in the figure below. All the [ birds are at the same height from the ground. Assume that the Eagle can fly at the same speed in all directions and that the geese are unaware of the Eagle's intention and will not move from their positions.

(Note: The figure is not to scale.)

If the eagle wants to attack the goose that is nearer to it, which one should it attack? Show your steps.

(Note: Use √2 = 1.41, √3 = 1.73)

Soln.  The eagle is located at the origin of the diagram.

The positions of the geese are:

• Goose 1: 100 m away at an angle of 45°.
• Goose 2: 75 m away at an angle of 30°.

Note: The distances given are along the hypotenuse of their respective right triangles

Distance to Goose 1

d_1​= √(100. cos(45°))² + (100. sin(45°))²

cos(45°)=sin(45°)= √2/2≈0.707

d₁​= √(100. 0.707)² + (100. 0.707)²

d_1​= √(70.7)² + (70.7)²

d_1​= √2 x (70.7)² 

d₁​= √2 x 5000

d₁​= √10000

d₁​= 100

Distance to Goose 2

d_2​= √(75. cos(30°))² + (75. sin(30°))²

d₂​= √(75. 0.866)² + (75. 0.5)²

d_2​= √(64.95)² + (37.5)²

d_2​= √4220.9 + 1406.25² 

d₂​= √5627.15

≈75m

Since d₂=75 is shorter than d₁=100 , the eagle should attack Goose 2, as it is closer.

4. In the giant wheel shown below, Gagan is sitting in one of the cabins which is 12 m high from the platform. Jyoti and Karan are sitting in the lowest and the highest cabins from the platform respectively.

(Note: The figure is not to scale.)

From Gagan, the angle of depression of Jothi and the angle of elevation of Kiran is 30° and 60° respectively.

i) What will be the angle of elevation of Gagan from Jothi?

ii) What will be the angle of depression of Gagan from Kiran?

iii) Find the diameter of the giant wheel.

Show your steps with a diagram.

Soln.  Let the height of the center of the wheel from the platform be h

Let the radius of the wheel be r, so h=r.

Gagan’s cabin is at 12 m, so h−12=r

Jyoti is at the bottom of the wheel, at height 0

Karan is at the top of the wheel, at height 2r

The angle of elevation from Jyoti to Gagan is 30°

Hence, the angle is complementary and equal to 30°

Gagan to Jyoti: From the geometry and the given angle of depression 30°, the relationship between the height h=12 and the radius r can be calculated.

​Radius of the wheel:

From Jyoti to Gagan, the radius of the wheel is approximately 20.78 m

Diameter of the wheel:

The diameter of the giant wheel is 41.57 m

Height of Karan:

The height of Karan (topmost cabin) is approximately 48.0 m

The angle of elevation from Jyoti to Gagan is the same as the angle of depression from Gagan to Jyoti, which is  30°.

The angle of depression of Gagan from Karan is the same as the angle of elevation of Karan from Gagan, which is 60°.

The diameter of the wheel is 41.57 m

5. A ship was moving towards the shore at a uniform speed of 36 km/h. Initially, the ship was 1.3 km away from the foot of a lighthouse which is 173.2 m in height.

Find the angle of depression, x, of the top of the lighthouse from the ship after the ship had been moving for 2 minutes.

Show your steps and give reasons.

(Note: Take √3 as 1.732 and √2 as 1.414.)

Soln. 

Height of the lighthouse = 173.2 m

Initial horizontal distance of the ship from the lighthouse = 1.3 km

Speed of the ship = 36 km

Time of travel = 2 minutes = 2/60 hours= 1/30hours

Distance = Speed ×Time

Distance traveled=36× 1/30

=1.2km.

The new horizontal distance = Initial distance −-− Distance traveled:

New distance=1.3−1.2

=0.1km.

tan(x)= Height of lighthouse/ New horizontal distance

= 0.1732​/ 0.1

tan(x)=1.732

Since tan⁡(x)=1.732, and we are given that √3=1.732,

x=60°

Some More Important Question Answers of Class 10 Maths Trigonometry 

Q1. The angle of elevation of the top of a tower from a point on the ground is 30°. If the point is 50 m away from the base of the tower, find the height of the tower.

Solution: Let the height of the tower = h m.
Distance from point to base = 50 m. 

Using tan θ = opposite/adjacent:
tan 30° = h / 50
1/√3 = h / 50
h = 50 / √3 ≈ 28.9 m

Answer: Height of tower = 28.9 m (approx).

Q2. From the top of a building 20 m high, the angle of depression of a car on the ground is 30°. Find the distance of the car from the building.

Solution: Height of building = 20 m.
Let distance of car from building = x m.

tan 30° = 20 / x
1/√3 = 20 / x
x = 20√3 ≈ 34.6 m

Answer: Distance = 34.6 m.

Q3. A kite is flying at a height of 60 m. The length of the string from the ground to the kite is 100 m. Find the angle of elevation of the kite from the ground (assume no sag).

Solution: Opposite = 60 m, Hypotenuse = 100 m.

sin θ = 60/100 = 0.6
θ = sin⁻¹(0.6) ≈ 36.9°

Answer: Angle of elevation ≈ 37°.

Q4. A ladder 15 m long rests against a wall. The foot of the ladder is 9 m away from the wall. Find the angle of elevation of the top of the ladder.

Solution: Opposite = 12 m, Adjacent = 9 m, Hypotenuse = 15 m.

tan θ = 12/9 = 4/3
θ = tan⁻¹(4/3) ≈ 53.1°

Answer: Angle = 53°.

Q5. A person is standing 50 m away from a flagpole. The angle of elevation of the top of the pole is 60°. Find the height of the pole.

Solution: tan 60° = h / 50
√3 = h / 50
h = 50√3 ≈ 86.6 m

Answer: Height = 86.6 m.

Q6. The angle of elevation of the top of a building from a point on the ground is 45°. On moving 20 m nearer to the building, the angle of elevation becomes 60°. Find the height of the building.

Solution: Let height = h, distance from building = x.

Case 1: tan 45° = h / x → h = x.
Case 2: tan 60° = h / (x – 20) → √3 = h / (x – 20).

Substitute h = x: √3 = x / (x – 20)
√3(x – 20) = x
√3x – 20√3 = x
(√3 – 1)x = 20√3
x = (20√3) / (√3 – 1).

Rationalize: x = (20√3)(√3 + 1) / (3 – 1)
x = (20√3)(√3 + 1) / 2 = 10√3(√3 + 1)
= 10(3 + √3) ≈ 43.2 m.

Height h = x = 43.2 m.

Q7. From the top of a building 60 m high, the angle of elevation of the top of a tower is 30° and the angle of depression of the base of the tower is 60°. Find the height of the tower.

Diagram hint: Two right triangles: one from top to base (depression), one from top to top (elevation).

Solution (Step-by-step): Let distance between building and tower = x.
Let tower height = H.

tan 60° = 60 / x → √3 = 60 / x → x = 60/√3 = 20√3.

Now for elevation: tan 30° = (H – 60) / x
1/√3 = (H – 60) / (20√3)
H – 60 = 20
H = 80 m.

Answer: Tower height = 80 m.

Q8. A balloon is tied to a rope fixed on the ground. The balloon is 100 m above the ground. If the length of the rope is 125 m, find the angle of elevation of the balloon.

Solution: sin θ = 100/125 = 0.8
θ = sin⁻¹(0.8) ≈ 53.1°

Answer: Angle = 53°.

Q9. The angle of elevation of the top of a hill from a point on the ground is 45°. After walking 100 m towards the hill, the angle of elevation becomes 60°. Find the height of the hill.

Solution: Let height = h, initial distance = x.

tan 45° = h / x → h = x.
tan 60° = h / (x – 100) → √3 = h / (x – 100).

Substitute h = x: √3 = x / (x – 100).

Solve: √3(x – 100) = x → √3x – 100√3 = x → (√3 – 1)x = 100√3.
x = (100√3) / (√3 – 1).

Approx: x ≈ 136.6 m → h = 136.6 m.

Answer: Height ≈ 136.6 m.

Q10. A man is watching a boat from the top of a lighthouse 40 m high. The angle of depression of the boat is 30°. Find the distance of the boat from the lighthouse.

Solution: tan 30° = 40 / x
1/√3 = 40 / x
x = 40√3 ≈ 69.3 m.

Answer: Distance ≈ 69 m.

Q11. A ladder 10 m long reaches a window which is 8 m above the ground. Find the angle of elevation of the top of the ladder.

Solution: sin θ = 8 / 10 = 0.8
θ = sin⁻¹(0.8) ≈ 53°.

Answer: Angle = 53°.

Q12. A man standing on a platform 20 m high observes a building whose angle of elevation is 30° and angle of depression of the base is 60°. Find the height of the building.

Solution: Let building height = H, distance between = x.

tan 60° = 20 / x → √3 = 20 / x → x = 20/√3.

tan 30° = (H – 20) / x → 1/√3 = (H – 20) / (20/√3).
H – 20 = 20 → H = 40 m.

Answer: Height = 40 m.

Q13. From a point on the ground, the angle of elevation of a bird on a tree is 45°. If the bird is 20 m above the ground, find the distance of the point from the foot of the tree.

Solution: tan 45° = 20 / x → 1 = 20 / x → x = 20 m.

Answer: Distance = 20 m.

Q14. The angle of elevation of the top of a building is 45°. If the building is 30 m high, find the distance of the observer from the building.

Solution: tan 45° = 30 / x → 1 = 30 / x → x = 30 m.

Answer: Distance = 30 m.

Q15. A boy standing 90 m away from a tower observes the angle of elevation of the top to be 30°. Find the height of the tower.

Solution: tan 30° = h / 90 → 1/√3 = h / 90 → h = 90/√3 ≈ 52 m.

Answer: Tower height = 52 m.

Q16. A ladder is placed against a wall. The top of the ladder touches the wall at a height of 24 m. If the ladder makes an angle of 60° with the ground, find the length of the ladder.

Solution: sin 60° = 24 / L → √3/2 = 24 / L → L = 24 × 2 / √3 = 48/√3 ≈ 27.7 m.

Answer: Ladder length ≈ 27.7 m.

Q17. A person observes the top of a temple at an angle of elevation of 60°. On moving 30 m away, the angle becomes 30°. Find the height of the temple.

Solution: Let temple height = h, initial distance = x.

tan 60° = h / x → √3 = h / x → h = √3x.
tan 30° = h / (x + 30) → 1/√3 = h / (x + 30).

Substitute h: 1/√3 = √3x / (x + 30).
x + 30 = 3x → 2x = 30 → x = 15.
h = √3 × 15 ≈ 25.98 m.

Answer: Height ≈ 26 m.

Q18. From the top of a 120 m tall building, the angle of elevation of the top of a tower is 30° and angle of depression of its base is 45°. Find the height of the tower.

Solution: Let tower height = H, distance between = x.

tan 45° = 120 / x → 1 = 120 / x → x = 120.
tan 30° = (H – 120) / 120 → 1/√3 = (H – 120) / 120.
H – 120 = 120/√3 ≈ 69.3.
H ≈ 189.3 m.

Answer: Tower height ≈ 189 m.

Q19. A man is standing on the bank of a river 60 m wide. He observes that the angle of elevation of the top of a tree on the opposite bank is 30°. Find the height of the tree.

Solution: tan 30° = h / 60 → 1/√3 = h / 60 → h = 60/√3 ≈ 34.6 m.

Answer: Height = 34.6 m.

Q20. The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving 40 m nearer to the tower, the angle of elevation becomes 60°. Find the height of the tower.

Solution: Let height = h, distance from A = x.

tan 30° = h / x → 1/√3 = h / x → h = x/√3.
tan 60° = h / (x – 40) → √3 = h / (x – 40).

Substitute: √3 = (x/√3) / (x – 40).
√3(x – 40) = x/√3 → 3(x – 40) = x → 3x – 120 = x → 2x = 120 → x = 60.
h = 60/√3 ≈ 34.6 m.

Answer: Tower height = 34.6 m.

Extra Questions for Practice!

Q1. A man observes the top of a minaret at an angle of elevation of 45°. If the minaret is 40 m tall, find his distance from it.

Q2. A kite flying at a height of 50 m is tied with a string of length 70 m. Find the angle of elevation of the kite.

Q3. A 15 m long ladder is placed against a wall. The foot is 9 m away from the wall. Find the angle the ladder makes with the ground.

Q4. The angle of depression of a boat from a lighthouse is 45°. If the height of the lighthouse is 40 m, find the distance of the boat.

Q5. From a point 25 m away from a tree, the angle of elevation of the top of the tree is 30°. Find the height of the tree.

Q6. A tower stands 100 m away from a building. The angle of elevation of the top of the tower from the roof of the building is 30°, and the angle of depression of the base is 60°. Find the height of the tower if the building is 20 m tall.

Q7. The angle of elevation of the top of a vertical pole is 60° when observed from a point A, and 30° from point B on the same line of sight. If AB = 100 m, find the height of the pole.

Q8. A balloon is observed at an angle of elevation of 45°. If the string attached to the balloon is 80 m, find the height of the balloon.

Q9. A boy standing at a distance of 24 m from a building observes its top at 30°. Find the height of the building.

Q10. The top of a tower is observed at 60° from one point, and 30° from another point in the same line, 100 m farther away. Find the height of the tower.

How to Ace These Class 10 Applications of Trigonometry Important Questions

This chapter is entirely application-based, so the key lies in understanding the problem and visualising the right triangle correctly. Follow this step-by-step method to prepare effectively:

Step 1: Revise Trigonometric Ratios from the Previous Chapter

Before solving applications, ensure that you remember all basic trigonometric ratios: sin θ, cos θ, tan θ, and their reciprocal relations. Most problems use tan θ because it relates perpendicular and base directly.

Step 2: Learn Key Terms – Angle of Elevation and Angle of Depression

• Angle of Elevation: The angle formed by the line of sight when looking up at an object from a horizontal line.
• Angle of Depression: The angle formed by the line of sight when looking down at an object from a horizontal line.

Always remember that these angles are measured from the horizontal line.

Step 3: Draw a Neat Diagram for Each Problem

Convert the question into a right triangle. Mark all known values (heights, distances, angles) and label the unknown quantities to be calculated. A clear diagram helps you choose the correct trigonometric ratio.

Step 4: Identify the Correct Trigonometric Ratio

• Use tan θ = Perpendicular / Base most often.
• For other situations, use sin θ = Perpendicular / Hypotenuse or cos θ = Base / Hypotenuse depending on what is given.

Step 5: Apply the Ratio and Solve Stepwise

Write the formula, substitute the given values, and simplify systematically. Always mention the trigonometric ratio used in the solution.

Step 6: Practise Word Problems with Two Angles

Some questions involve two angles of elevation or depression from different points. Draw separate right triangles for each and use the data to form equations step by step. It also helps to revise the concepts from the NCERT Book Class 10 Maths before practising questions.

Step 7: Remember Practical Conditions

Unless stated otherwise, keep in mind that the observer’s eye level or the base of objects like towers and buildings are on the same horizontal line.

How to Make the Best Use of Trigonometry Class 10 Important Questions?

Now that we understand the importance of practicing trigonometry in class with 10 important questions, let’s learn how it can benefits you to maximize your learning.

• Real-Life Relevance: Shows how mathematics applies to everyday measurement problems.
• Better Visualisation: Improves the ability to draw and interpret geometric situations.
• High Scoring Chapter: Straightforward applications ensure accuracy with correct formulas.
• Strengthened Trigonometric Understanding: Reinforces sin, cos, and tan concepts from the previous chapter.‍
• Logical Problem Solving: Develops analytical thinking by converting word problems into equations.

FAQs

Q1. How many marks are generally allotted to this chapter in Class 10 board exams?

Ans. This chapter usually carries 3 to 5 marks, generally through one or two word problems.

Q2. Which trigonometric ratio is used most often?

Ans. tan θ is most commonly used because it directly relates height and base in right triangles.

Q3. How can I identify which ratio to use in a question?

Ans. Look at what quantities are given.

• If perpendicular and base are involved, use tan θ.
• If perpendicular and hypotenuse are involved, use sin θ.
• If base and hypotenuse are involved, use cos θ.

Q4. Are diagrams compulsory for these questions?

Ans. Yes, diagrams are essential for full marks. They help you visualise the problem and show the examiner your understanding of geometry.

Q5. How can I avoid confusion between angle of elevation and depression?

Ans. Remember: elevation means looking up, and depression means looking down from a horizontal line. Mark them clearly on your diagram.