Some Applications of Trigonometry: Important Questions and Answers for Class 10

Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry are available in this article to help students prepare effectively for CBSE board exams. These questions are designed according to the latest CBSE exam pattern and revised syllabus, making them highly useful for exam-oriented practice. The chapter Some Applications of Trigonometry is an important part of Class 10 Maths because it explains the practical use of trigonometry in finding heights and distances. Questions from this chapter are frequently asked in board exams as well as competitive exams like JEE and other entrance tests. Practising these questions with solutions will help students build confidence, enhance accuracy, and score better marks in the examination.

Table of Contents

• Key Concepts of Applications of Trigonometry for Class 10
• Important Questions on Applications of Trigonometry for Class 10

Key Concepts of Applications of Trigonometry for Class 10

The chapter introduces three core ideas that every question revolves around:

• Angle of Elevation: Angle of elevation is the angle formed between a horizontal line and the line of sight when you look upward at an object.

• Angle of Depression: The angle formed when you look downward from the horizontal to see an object below you, like a pilot looking down at a runway.

• Line of Sight: The straight imaginary line drawn from your eye to the object you are observing.

Essential Trigonometry Values:

Important Questions on Applications of Trigonometry for Class 10

Q1: 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°. (2 Marks)

Solution: Given, length of rope = 20 m

sin 30° = AB / AC = AB / 20

Substitute sin 30° = 1/2.

1/2 = AB / 20 

⇒ AB = 20 × (1/2) = 10 m

The height of the pole is 10 metres.

Q2: A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground, making an angle of 30° with it. The distance from the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. (2 Marks)

Solution: ∠ACB = 30°.

cos 30° = BC/AC. 

⇒ √3/2 = 8/AC 

⇒ AC = 16/√3

tan 30° = AB/BC 

⇒ 1/√3 = AB/8 

⇒ AB = 8/√3

Total height = AB + AC (standing part + broken part):

= 8/√3 + 16/√3 = 24/√3 = 24√3/3 = 8√3 m

The original height of the tree was 8√3 metres ≈ 13.86 m.

Q3: The shadow of a tower standing on level ground is found to be 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the height of the tower. (3 Marks)

Solution: Let the tower height = h metres. 

When the sun's altitude (angle of elevation of the sun) is 60°, the shadow length = x m. 

When it is 30°, the shadow = (x + 40) m.

For the triangle at 60°:

tan 60° = h/x 

⇒ √3 = h/x 

⇒ h = x√3 ... (i)

For the triangle at 30°:

tan 30° = h/(x + 40) 

⇒ 1/√3 = h/(x+40) ... (ii)

Substitute (i) into (ii):

1/√3 = x√3/(x+40) 

⇒ x+40 = 3x

⇒ 2x = 40 

⇒ x = 20

From (i): h = 20√3 m

The height of the tower is 20√3 metres ≈ 34.64 m.

Q4: Two poles of equal heights are standing opposite each other on either side of a road that is 80 m wide. From a point between them on the road, the angles of elevation of the tops of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles. (3 Marks)

Solution: Let AB and CD be equal poles. 

Point O lies on the road between them. 

OB + OD = 80 m. 

Let OB = d, so OD = (80 – d). 

Let height = h m.

From O toward pole AB (angle 60°):

tan 60° = h/OB 

⇒ √3 = h/d 

⇒ h = d√3 ... (i)

From O toward pole CD (angle 30°):

tan 30° = h/OD 

⇒ 1/√3 = h/(80-d) 

⇒ h = (80-d)/√3 ... (ii)

Equate (i) and (ii):

d√3 = (80 - d)/√3 

⇒ 3d = 80-d 

⇒ 4d = 80 

⇒ d = 20 m

So OD = 80 – 20 = 60 m. From (i): h = 20√3 m

Height of each pole = 20√3 m. 

The distance from the point to the nearer pole is 20 m and to the farther pole is 60 m.

Q5: An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer. (3 Marks)

Solution: The observer's eye is at a 1.5 m height. The tower is 22 m tall.

The effective height difference = 22 – 1.5 = 20.5 m.

The right triangle has:

Opposite side (vertical) = 20.5 m

Adjacent side (horizontal) = 20.5 m

tan θ = 20.5 / 20.5 = 1 

⇒ θ = 45°

The angle of elevation is 45°.

Q6: 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 cable tower. (5 Marks)

Solution: Building height = 7 m. 

Let the horizontal distance between them = d m. Let the cable tower height = h m.

Using the angle of depression to the foot of the tower (45°): the horizontal distance d equals the height of the building (since tan 45° = 1).

tan 45° = 7/d 

⇒ 1 = 7/d 

⇒ d = 7 m

The top of the cable tower is (h – 7) m above the observer's eye level. Using angle of elevation (60°):

tan 60° = (h – 7)/d 

⇒ √3 = (h – 7)/7

⇒ h – 7 = 7√3 

⇒ h = 7 + 7√3 = 7(1 + √3) m

Height of the cable tower = 7(1 + √3) m ≈ 19.12 m.

Q7: The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 20 m towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower. (5 Marks)

Solution: Original angle = 30°. 

After moving 20 m closer, the new angle = 30° + 15° = 45°. 

Let height = h, and the new (closer) distance = x.

From the closer point (angle = 45°):

tan 45° = h/x 

⇒ 1 = h/x 

⇒ x = h ... (i)

From the original point (angle = 30°), the distance is (x + 20):

tan 30° = h/(x + 20) 

⇒ 1/√3 = h/(x+20) ... (ii)

Substituting x = h from (i) into (ii):

(h + 20) = h√3 

⇒ h(√3 – 1) = 20

Rationalise:

h = 20/(√3–1) × (√3+1)/(√3+1) = 20(√3+1)/2 = 10(√3+1) m

The height of the tower is 10(√3 + 1) m ≈ 27.32 m.

Q8: The angles of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √(st). (3 Marks)

Solution: Let the tower height = h. Points B and C are at distances s and t from the foot. A. Since the angles are complementary, if ∠ABT = θ, then ∠ACT = (90° – θ).

From point B (distance s):

tan θ = h/s ... (i)

From point C (distance t):

tan(90°–θ) = h/t 

⇒ cot θ = h/t ... (ii)

Multiply equations  (i) and (ii):

tan θ × cot θ = (h/s) × (h/t) 

⇒ 1 = h²/(st)

h² = st 

⇒ h = √(st) (since h > 0)

Hence proved: Height of the tower = √(st).

Q9: 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. (5 Marks)

Solution: Lighthouse height = 75 m. The two ships S₁ (closer) and S₂ (farther) are on the same side. 

Angle of depression to S₁ = 45°, to S₂ = 30°. 

Let the distance of S₁ from the base = d₁ and S₂ = d₂.

For closer ship S₁ (angle of depression = 45°, using alternate angles: angle of elevation from S₁ = 45°):

tan 45° = 75/d₁ 

⇒ 1 = 75/d₁ 

⇒ d₁ = 75 m

For farther ship S₂ (angle of depression = 30°):

tan 30° = 75/d₂ 

⇒ 1/√3 = 75/d₂ 

⇒ d₂ = 75√3 m

Distance between the ships:

= d₂ – d₁ = 75√3 – 75 = 75(√3 – 1) m

The distance between the two ships = 75(√3 – 1) m ≈ 54.9 m.

Q10: 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 30 seconds, the angle of elevation reduces to 30°.

Based on the above information, answer the following questions. (Take √3 =1.732)

(i) Find the distance travelled by the balloon during the interval. (2)

(ii) Find the speed of the balloon. (2)

Solution: (i) In the figure, let C be the position of the observer (the girl).

A and P are two positions of the balloon.

CD is the horizontal line from the eyes of the (observer) girl.

Here PD = AB = 88.2 m − 1.2 m = 87 m

In ∆ ABC, we have AB/BC = tan 60° 

⇒ 87/BC = √3

⇒ BC = 87/√3 m

In ∆ PDC, PD/CD = tan 30°

⇒ 87/CD = 1/√3 ⇒ CD = 87√3 cm

Now BD = CD - BC = 87√3 - 87/√3 = 58√3 m 

Thus, the required distance between the two positions of the balloon = 58 √3 m

= 58 x 1.732 = 100.46 m (approx.)

(ii) Speed of the balloon = Distance/time = 100.46/30 = 3.35 m/s (approx.)

Q11: If the angle of elevation of the sun is 60°, the length of the shadow of a 30 m tall tree is:

Solution: tan 60° = 30/shadow 

⇒ √3 = 30/shadow 

⇒ shadow = 30/√3 = 10√3 m

Answer: 10√3 m

Q 12: A pole 6 m high casts a shadow 6√3 m long. The angle of elevation of the sun is:

Solution: tan θ = 6/(6√3) = 1/√3 

⇒ θ = 30°

Answer: 30°

Q13: From a point on the ground, the angle of elevation of the top of a 10 m tall tower is 45°. The distance of the point from the base of the tower is:

Solution: tan 45° = 10/d 

⇒ 1 = 10/d 

⇒ d = 10 m

Answer: 10 m

Learn more: Application Problems Trigonometry