Case Study Questions on Chapter 9 'Some Applications of Trigonometry' for Class 10 with Answers

This collection of case-study questions for Chapter 9: Some Applications of Trigonometry Class 10 follows CBSE and NCERT guidelines and presents short, exam-style scenarios designed to strengthen spatial reasoning, modelling, and step-by-step problem solving. Problems focus on interpreting translating real-life contexts into solvable trigonometric models. The set includes identification of appropriate trigonometric relations, derivation and use of ratio-based equations, multi-step problems requiring algebraic manipulation, and tasks that combine trigonometry with Pythagoras’ theorem. Answers are provided with clear reasoning and concise steps to help learners build procedural fluency and exam confidence. The downloadable PDF includes additional practice sets for quick revision and classroom worksheets to improve speed and accuracy.

Solved Some Applications of Trigonometry Case Study Questions and Answers

CBSE's case-based questions usually give you a short passage describing a real situation, sometimes with a small figure, followed by 4 - 5 questions of increasing difficulty, a couple of MCQs to check basic understanding, a short-answer question that needs a calculation, and a longer question that asks you to interpret or justify your answer.

Case Study 1:The Drifting Balloon

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 a certain instant, is 60°. After 30 seconds, the angle of elevation reduces to 30°. Take √3 = 1.732.

(i) The height of the balloon above the girl's eye level is:

(a) 88.2 m

(b) 87 m

(c) 89.4 m

(d) 1.2 m

(ii) Find the distance travelled by the balloon during the 30-second interval.

(iii) Fill in the Blank: The speed of the balloon during this interval is ______ m/s (approx).

(iv) True/False: If instead the elevation of the sun is 30°, a 150-foot tower would cast a shadow longer than the tower itself.

Solution: 

Subtract the girl's eye level from the balloon's height: 88.2 − 1.2 = 87 m. This 87 m becomes the vertical side of both right triangles you'll draw. Answer: (b) 87 m

(ii)  Let the girl's eye level be C, and let A, P be the two balloon positions, with CD the horizontal line of sight.

In the first triangle: tan 60° = 87/CB 

⇒ CB = 87/√3 = 29√3 m

In the second triangle: tan 30° = 87/CD 

⇒ CD = 87√3 m

Distance travelled = CD − CB = 87√3 − 29√3 = 58√3 m = 58 × 1.732 ≈ 100.46 m

(iii) Speed = Distance ÷ Time = 100.46 ÷ 30 ≈ 3.35 m/s

(iv) tan 30° = 150/BC 

⇒ BC = 150√3 feet ≈ 259.8 feet, which is indeed longer than 150 feet. 

Answer: True

Case Study 2: The Fire Brigade Ladder

A fire brigade van is parked 6 m away from a building. The fireman places a ladder against the building from the van so that its top reaches a window 8 m above the ground. The ladder makes an angle θ with the ground.

(i) Find the length of the ladder.

(ii) The angle θ that the ladder makes with the ground is:

(a) 30°

(b) 45°

(c) tan⁻¹(4/3)

(d) 60°

(iii) Fill in the Blank

If the van had been parked only 4 m from the building and the window height stayed at 8 m, the ladder would need to be at least ______ m long.

Solution: 

(i) The ladder is the hypotenuse of a right triangle with base 6 m and height 8 m. 

By Pythagoras: ladder² = 6² + 8² = 36 + 64 = 100 

⇒ ladder = 10 m

(ii) tan θ = 8/6 = 4/3, which is not one of the standard 30°–45°–60° angles, so θ = tan⁻¹(4/3) ≈ 53.13°. 

Answer: (c) tan⁻¹(4/3)

(iii) ladder² = 4² + 8² = 16 + 64 = 80 

⇒ ladder = √80 = 4√5 ≈ 8.94 m

Case Study 3: Two Ships and a Lighthouse

As observed from the top of a 100 m high lighthouse from sea level, the angles of depression of two ships are 30° and 45°. One ship is exactly behind the other on the same side of the lighthouse. Take √3 = 1.732.

(i) The distance of the nearer ship (45° depression) from the base of the lighthouse is:

(a) 100 m

(b) 100√3 m

(c) 50 m

(d) 173.2 m

(ii) Find the distance of the farther ship (30° depression) from the base of the lighthouse.

(iii) Fill in the Blank

 The distance between the two ships is ______ m.

(iv) True/False

The angle of depression from the top of the lighthouse to the nearer ship is greater than to the farther ship.

Solutions:

(i) By alternate angles, the angle of elevation from the ship equals the depression = 45°.

 tan 45° = 100/d 

⇒ d = 100 m. 

Answer: (a) 100 m

(ii) tan 30° = 100/d 

⇒ d = 100√3 = 100 × 1.732 = 173.2 m

(iii) Distance between ships = 173.2 − 100 = 73.2 m

True/False

(iv) Yes, the closer an object is, the steeper (larger) the angle of depression needed to see it from the same height. 45° > 30°. 

Answer: True

Case Study 4:  Kite Flying on Republic Day

On Republic Day, a group of students flies a kite from level ground using a string. The string makes an angle of 60° with the ground, and the length of string let out is 40 m. Assume the string is taut and straight.

(i) The height of the kite above the ground is:

(a) 20 m

(b) 20√3 m

(c) 40√3 m

(d) 40/√3 m

(ii) Find the horizontal distance of the kite from the student.

(iii) Fill in the Blank

 If the string is let out further so that the kite reaches a height of 20 m while the angle stays at 60°, the new string length used so far is ______ m.

Solution: 

sin 60° = height/40 

⇒ height = 40 × (√3/2) = 20√3 m. 

Answer: (b) 20√3 m

(ii) cos 60° = distance/40 

⇒ distance = 40 × (1/2) = 20 m

Fill in the Blank

(iii) sin 60° = 20/L 

⇒ L = 20 ÷ (√3/2) = 40/√3 = (40√3)/3 ≈ 23.09 m

Click below to download your free Case Study 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 Case Study PDF