HOTS Questions on Class 10 Maths Chapter 9 ‘Some Applications of Trigonometry’ with Answers

Higher-Order Thinking Skills (HOTS) question set for Chapter 9 ‘Some Applications of Trigonometry’ for Class 10 is crafted to move students beyond routine calculation and nurture transferable problem-solving skills using trigonometric methods. Aligned with CBSE and NCERT learning outcomes, these questions emphasise conceptual understanding and application of ideas such as heights and distances, using angle of elevation and depression, applying trigonometric ratios and solving multi-step real-world problems that require careful assumption, approximation, and interpretation. Each question includes a clear answer and brief reasoning to support classroom discussion and self-study. Download the PDF to access the same HOTS questions in a printable format, making offline learning and exam revision easier.

Solved HOTS Questions on Some Applications of Trigonometry

Question 1: The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 30 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower and the original distance from the tower.

Solution: 

tan 60° = h/d ⟹ h = d√3  ... (1)

From A (angle 30°), AQ = d + 30:

tan 30° = h/(d+30) 

⟹ h = (d+30)/√3  ... (2)

Equate (1) and (2):

d√3 = (d+30)/√3 

3d = d + 30 

2d = 30 

d = 15 m

Substitute into (1):

h = 15√3 m

Original distance AQ = d + 30 = 15 + 30 = 45 m

Height = 15√3 m ≈ 25.98 m and Original distance = 45 m

Question 2: A flagstaff of height 5 m stands on top of a tower. From a point on the ground, the angle of elevation of the top of the flagstaff is 60° and the angle of elevation of the top of the tower is 45°. Find the height of the tower.

Solution: 

Angle of elevation to top of tower (B) = 45°:

tan 45° = h/d 

⟹ 1 = h/d 

⟹ d = h ... (1)

Angle of elevation to top of flagstaff (C) = 60°:

tan 60° = (h+5)/d 

⟹ √3 = (h+5)/d ... (2)

Substitute d = h into (2):

√3 = (h+5)/h 

√3 h = h + 5 

h(√3 − 1) = 5 

h = 5/(√3 − 1)

Rationalise:

h = 5(√3 + 1)/((√3 − 1)(√3 + 1)) = 5(√3 + 1)/(3 − 1) = 5(√3 + 1)/2 m

Height of tower = 5(√3 + 1)/2 m ≈ 6.83 m

Question 3: From the top of a cliff 80 m high, the angles of depression of two boats on the same side are 45° and 30°. Find the distance between the two boats.

Solution: 

For the nearer boat C (angle of elevation = 45°):

tan 45° = 80/BC 

⟹ 1 = 80/BC 

⟹ BC = 80 m

For the farther boat D (angle of elevation = 30°):

tan 30° = 80/BD 

⟹ 1/√3 = 80/BD 

⟹ BD = 80√3 m

Distance between the two boats:

CD = BD − BC = 80√3 − 80 = 80(√3 − 1) m

Distance between boats = 80(√3 − 1) m ≈ 58.56 m

Question 4: From the top of a building 60 m high, the angles of elevation and depression of the top and foot of another building on the opposite side of the road are 60° and 45° respectively. Find the height of the other building and the width of the road.

Solution: 

Angle of depression to D (from A) = 45°. 

In △ABD, right-angled at B:

tan 45° = AB/BD = 60/d 

1 = 60/d 

⟹ d = 60 m

Angle of elevation to C (from A) = 60°. 

In △ACE, where CE = H − 60 and AE = d = 60:

tan 60° = CE/AE = (H−60)/60 

√3 = (H−60)/60 

H − 60 = 60√3 

H = 60 + 60√3 = 60(1 + √3) m

Width of road = 60 m and height of other building = 60(1 + √3) m ≈ 163.9 m

Question 5: Two buildings are on opposite sides of a road. From the top of the first building of height 10 m, the angle of elevation of the top of the second building is 60°. From the same top, the angle of depression of the foot of the second building is 30°. Find the height of the second building and the width of the road.

Solution:

Angle of depression to D = 30°. In △ABD:

tan 30° = AB/BD = 10/d 

1/√3 = 10/d ⟹ d = 10√3 m

Angle of elevation to C = 60°. In △ACE, AE = d = 10√3:

tan 60° = CE/AE = (H−10)/10√3 

√3 = (H−10)/10√3

H − 10 = 10√3 × √3 = 30 H = 40 m

Height of second building = 40 m and Width of road = 10√3 m ≈ 17.3 m 

Question 6: A vertical pole of height 6 m casts a shadow 4 m long on the ground. At the same time, a tower casts a shadow 28 m long. Find the height of the tower.

Solution: 

Since both shadows are cast at the same time, the sun's angle of elevation θ is identical for both.

For the pole:

tan θ = 6/4 = 3/2

For the tower (height H, shadow 28 m):

tan θ = H/28

Since tan θ is the same:

H/28 = 3/2 H = 28 × 3/2 = 42 m

Height of the tower = 42 m 

Question 7: As observed from the top of a lighthouse, 75 m high above sea level, the angles of depression of two ships on opposite sides of the lighthouse are 30° and 45°. Find the distance between the two ships.

Solution: 

For Ship C (angle of elevation from C = 30°):

tan 30° = 75/BC 

⟹ BC = 75√3 m

For Ship D (angle of elevation from D = 45°):

tan 45° = 75/BD 

⟹ BD = 75 m

Distance between ships:

CD = BC + BD = 75√3 + 75 = 75(√3 + 1) m

Distance between the two ships = 75(√3 + 1) m ≈ 204.9 m

 

Questions PDF with worked-out examples for Class 10 Chapter 9: Some Applications of Trigonometry, perfect for last-minute CBSE exam revision.

Class 10 Chapter 9: Some Applications of Trigonometry HOTS PDF