Class 10 Maths MCQs for Chapter 9 Some Applications of Trigonometry with Answers PDF

MCQs on Some Applications of Trigonometry Class 10 are available in this Maths article along with a free PDF for offline practice. These multiple‑choice questions help students practise the key concepts from Chapter 9 of the CBSE Maths syllabus in an exam‑oriented format. The MCQs with answers and detailed solutions, prepared by our subject experts, cover angle of elevation, angle of depression, line of sight, horizontal level, and height‑and‑distance problems to strengthen conceptual understanding and improve problem‑solving skills. By practising MCQs on Some Applications of Trigonometry, students can improve accuracy, understand real‑life applications of trigonometry, and build confidence for board exams.

MCQS on Chapter 9: Some Applications of Trigonometry for Class 10 With Answers

Question 1: The angle formed by the line of sight with the horizontal when the point being viewed is ABOVE the horizontal level is called:

(a) Angle of elevation

(b) Angle of depression

(c) Angle of incidence

(d) None of the above

Answer: (a) Angle of elevation

Explanation: When we raise our head to look at an object above the horizontal level, the angle formed is the angle of elevation.

Question 2: If the length of the shadow of a tree is decreasing, the angle of elevation of the Sun is:

(a) Increasing

(b) Decreasing

(c) Remains the same

(d) Becomes zero

Answer: (a) Increasing

Explanation: As the shadow shortens, the observer's viewpoint moves closer to the base of the tree. With the height fixed, tan(θ) = height/shadow length increases as shadow length decreases. So θ increases.

Question 3: If both the height of a building and the distance from the building to an observation point are increased by 20%, the angle of elevation:

(a) Increases

(b) Decreases

(c) Does not change

(d) Becomes 90°

Answer: (c) Does not change

Explanation: tan θ = height/distance. If both are multiplied by 1.2, the ratio remains the same, so θ stays unchanged.

Question 4: The angle of elevation of the top of a building from a point 30 m away from its foot is 30°. The height of the building is:

(a) 10 m

(b) 30/√3 m = 10√3 m

(c) 30 m

(d) √3/10 m

Answer: (b) 30/√3 m = 10√3 m

tan 30° = h/30 ⇒ 1/√3 = h/30 ⇒ h = 30/√3 = 10√3 m.

Question 5: A kite is flying at a height of 60 m above the ground. The string makes an angle of 60° with the horizontal. The length of the string (assuming no slack) is:

(a) 40√3 m

(b) 40√3 m

(c) 60√3 m

(d) 120/√3 m

Answer: 40√3 m

Explanation: sin 60° = height/string length ⇒ √3/2 = 60/L ⇒ L = 60 × 2/√3 = 120/√3 = 40√3 m. 

Question 6: A tower 6 m high casts a shadow 2√3 m long on the ground. The angle of elevation of the Sun is:

(a) 60°

(b) 45°

(c) 30°

(d) 90°

Answer: (a) 60°

Explanation: tan θ = height/shadow = 6/(2√3) = 3/√3 = √3 = tan 60°. So θ = 60°.

Question 7: A circus artist climbs a 20 m long rope tied from the top of a vertical pole to the ground. If the rope makes a 30° angle with the ground, the height of the pole is:

(a) 20 m

(b) 20√3 m

(c) 10 m

(d) 10√3 m

Answer: (c) 10 m

Explanation: sin 30° = height/rope ⇒ 1/2 = h/20 ⇒ h = 10 m.

Question 8: From a point on a bridge at a height of 3 m from the banks, the angles of depression of the two banks are 30° and 45°. The width of the river is:

(a) 3(1 + √3) m

(b) 3 + √3 m

(c) 3(√3 + 1) m

(d) √3 + 3 m

Answer: (c) 3(√3 + 1) m

Explanation: For a 30° depression, tan 30° = 3/AD → AD = 3√3 m. For 45° depression: BD = DP = 3 m.

Width AB = AD + BD = 3√3 + 3 = 3(√3 + 1) m. 

Question 9: From the top of a 75 m high lighthouse, the angles of depression of two ships on the same side are 30° and 45°. The distance between the two ships is:

(a) 75(√3 − 1) m

(b) 75(√3 − 1) m

(c) 75√3 m

(d) 75 m

Answer: 75(√3 − 1) m

Explanation: Ship 1 (depression 45°): distance = 75 m. Ship 2 (depression 30°): distance = 75√3 m.

Gap between ships = 75√3 − 75 = 75(√3 − 1) m.

Question 10: Two poles of equal height stand opposite each other on a road 80 m wide. From a point between them, angles of elevation are 60° and 30°. The height of each pole is:

(a) 10√3 m

(b) 20√3 m

(c) 20√3 m

(d) 40√3 m

Answer: 20√3 m

Explanation: Let the point be at a distance x from the pole with 60° elevation. Then tan 60° = h/x ⇒ h = x√3. tan 30° = h/(80−x) ⇒ h = (80−x)/√3.

x√3 = (80−x)/√3 ⇒ 3x = 80−x ⇒ x = 20. h = 20√3 = 20√3 m.

Question 11: The angle of elevation of the top of a building from the foot of a tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, the height of the building is:

(a) 50/3 m

(b) 50/3 m

(c) 25 m

(d) 50√3 m

Answer: 50/3 m

Explanation: Let d = distance between them. From tower foot: tan 60° = 50/d ⇒ d = 50/√3.

From building foot: tan 30° = h/d ⇒ h = d/√3 = (50/√3)/√3 = 50/3 m.

Question 12: The angle of elevation of the top of a building from two points P and Q at distances 9 m and 4 m, respectively, from the base (in a straight line) are complementary. The height of the building is:

(a) √13 m

(b) 6 m

(c) 36 m

(d) √36 m

Answer: (b) 6 m

Explanation: Let height = h. tan θ = h/9, and tan(90°−θ) = cot θ = h/4.

So tan θ × cot θ = (h/9)(h/4) = 1 ⇒ h² = 36 ⇒ h = 6 m.

Question 13: The angle of elevation of the top of a building from two points 'p' and 'q' metres from the base (on the same side) are α and β, where α + β = 90°. The height of the building is:

(a) (p + q)/2

(b) pq

(c) √(pq)

(d) (p − q)/2

Answer: (c) √(pq)

Explanation: tan α = h/p and tan β = h/q. Since β = 90° − α, tan β = cot α = 1/tan α.

So h/q = 1/(h/p) = p/h ⇒ h² = pq ⇒ h = √(pq).

Question 14: The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving 20 m toward the tower, the elevation becomes 60°. The height of the tower is:

(a) 10 m

(b) 10√3 m

(c) 20 m

(d) 20√3 m

Answer: (b) 10√3 m

Explanation: Let height = h, original distance = d. tan 30° = h/d  ⇒ d = h√3. tan 60° = h/(d−20)  ⇒ h = (h√3 − 20)√3 = 3h − 20√3.

 ⇒ 2h = 20√3  ⇒ h = 10√3 m.

Question 15: If the height of the tower and the length of its shadow are equal, the angle of elevation of the Sun is:

(a) 30°

(b) 60°

(c) 45°

(d) 90°

Answer: (c) 45°

Explanation: tan θ = height/shadow = h/h = 1 = tan 45°. So θ = 45°.

 

Click here to download the free PDF of MCQs worksheet on Chapter 9: Some Applications of Trigonometry for Class 10 Maths based on the updated NCERT & CBSE pattern with important multiple-choice questions and answers.

MCQs Worksheet on Chapter 9: Some Applications of Trigonometry for Class 10