CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 11

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 09 Some Applications of Trigonometry

Explore reliable objective questions for Chapter 09 Some Applications of Trigonometry tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Practice Chapter 09 Some Applications of Trigonometry MCQs for Class 10 Mathematics

Access the complete set of multiple-choice questions for Chapter 09 Some Applications of Trigonometry below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: From the top of a tower 120 m high, the angle of depression of a car on level ground is 30°. The horizontal distance of the car from the foot of the tower is:
(a) \( 40\sqrt3 \) m
(b) 120 m
(c) \( 60\sqrt3 \) m
(d) \( 120\sqrt3 \) m (about 207.84 m)
Show Answer & Explanation

Answer: (d) \( 120\sqrt3 \) m (about 207.84 m)

Explanation:
1. tan 30° = 120/x, so x = 120/tan 30° = 120√3 ≈ 207.84 m.

Teacher's Note:
1. The angle at the car equals the angle of depression, by alternate angles.
2. Use tan = opposite/adjacent.

Question 2: The top of a hill is seen at an elevation of 60° from the foot of a tower, and the top of the tower (50 m high) is seen at an elevation of 30° from the foot of the hill. The tower and hill stand on level ground. The height of the hill is:
(a) \( 50\sqrt3 \) m
(b) 100 m
(c) 150 m
(d) 75 m
Show Answer & Explanation

Answer: (c) 150 m

Explanation:
1. Distance between the feet x = 50/tan 30° = 50√3 m. Hill height = x tan 60° = 50√3 × √3 = 150 m.

Teacher's Note:
1. First find the distance between the two feet from the tower triangle.
2. Reuse that distance in the hill triangle.

Question 3: Ranjan sits 8 m above the ground on a tree on a small island in the middle of a river, directly in line with the foot of the tree and two poles on opposite banks. The angles of depression of the feet of the poles are 60° and 30°. The width of the river is about:
(a) 9.24 m
(b) 18.475 m
(c) 27.7 m
(d) 13.86 m
Show Answer & Explanation

Answer: (b) 18.475 m

Explanation:
1. Distances from the tree foot: 8/tan 60° = 8/√3 and 8/tan 30° = 8√3. Width = 8/√3 + 8√3 = 32/√3 ≈ 18.475 m.

Teacher's Note:
1. Each bank is at a different horizontal distance from the foot of the tree.
2. Add the two distances.

Question 4: From the top of a lighthouse 100 m above sea level, the angle of depression of a ship sailing straight towards it changes from 30° to 45°. The distance travelled by the ship in this time is about:
(a) 73.2 m
(b) 100 m
(c) 173.2 m
(d) 63.4 m
Show Answer & Explanation

Answer: (a) 73.2 m

Explanation:
1. Distances from the base: 100/tan 30° = 173.2 m and 100/tan 45° = 100 m. Difference = 73.2 m.

Teacher's Note:
1. Find the ship's horizontal distance at each position.
2. Subtract.

Question 5: The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of Q is 45°. The height PQ of the tower is:
(a) 54.64 m
(b) 109.3 m
(c) 80 m
(d) \( 20(\sqrt3+1)+40 \) m ≈ 94.64 m
Show Answer & Explanation

Answer: (d) \( 20(\sqrt3+1)+40 \) m ≈ 94.64 m

Explanation:
1. Let the horizontal distance be x. From Y: the height of Q above Y is x tan 45° = x, so PQ = x + 40. From X: PQ = x tan 60° = √3 x. Hence √3 x = x + 40, x = 20(√3 + 1) ≈ 54.64 and PQ ≈ 94.64 m.

Teacher's Note:
1. The horizontal distance is the same from X and Y.
2. Write PQ in two ways and equate.

Question 6: Two pillars of equal height stand on either side of a road 100 m wide. The angles of elevation of their tops from a point P on the road between them are 60° and 30°. The distance of P from the pillar that makes 60° is:
(a) 75 m
(b) 25 m
(c) 50 m
(d) 20 m
Show Answer & Explanation

Answer: (b) 25 m

Explanation:
1. Let that distance be x: h = x tan 60° = √3 x, and h = (100 − x) tan 30° = (100 − x)/√3. So 3x = 100 − x, x = 25 m (h = 25√3 ≈ 43.3 m).

Teacher's Note:
1. Both pillars have the same height h.
2. Equate the two expressions for h.

Question 7: From a window 15 m above the ground, the angle of elevation of the top of the house opposite is 30° and the angle of depression of its foot is 45°. The height of the opposite house is about:
(a) 23.66 m
(b) 20.77 m
(c) 30 m
(d) 15 m
Show Answer & Explanation

Answer: (a) 23.66 m

Explanation:
1. Horizontal distance = 15/tan 45° = 15 m. Height above the window level = 15 tan 30° = 5√3 ≈ 8.66 m. Total = 15 + 8.66 = 23.66 m.

Teacher's Note:
1. The foot of the house is 15 m below the window.
2. Add the part above window level.

Question 8: From the top of a building 60 m high, the angles of depression of the top and the bottom of a tower are 30° and 60°. The height of the tower is:
(a) 20 m
(b) 30 m
(c) 60 m
(d) 40 m
Show Answer & Explanation

Answer: (d) 40 m

Explanation:
1. Distance between them x = 60/tan 60° = 20√3 m. Drop to the top of the tower = x tan 30° = 20 m. Height = 60 − 20 = 40 m.

Teacher's Note:
1. The horizontal distance is fixed by the angle to the bottom.
2. Subtract the part of the building above the tower top.

Question 9: A man on a ship's deck, 10 m above the water, sees the top of a hill at an elevation of 60° and its base at a depression of 30°. The height of the hill is:
(a) 30 m
(b) 10√3 m
(c) 40 m
(d) 50 m
Show Answer & Explanation

Answer: (c) 40 m

Explanation:
1. Distance = 10/tan 30° = 10√3 m. Part above deck level = 10√3 × tan 60° = 30 m. Hill height = 30 + 10 = 40 m.

Teacher's Note:
1. The base of the hill is at water level, 10 m below the deck.
2. Add the 10 m below the observer.

Question 10: The slope of a hill makes an angle of 60° with the horizontal. To reach the top one has to walk 500 m along the slope. The height of the hill is:
(a) 250 m
(b) \( 500\sqrt3 \) m
(c) \( \frac{500}{\sqrt3} \) m
(d) \( 250\sqrt3 \) m
Show Answer & Explanation

Answer: (d) \( 250\sqrt3 \) m

Explanation:
1. Height = 500 sin 60° = 500 × √3/2 = 250√3 m.

Teacher's Note:
1. The slope length is the hypotenuse.
2. Use sin.

Question 11: A hill slope makes 60° with the horizontal and one has to walk 300 m along the slope to reach the top. The horizontal distance from the foot of the slope to the point below the top is:
(a) \( 150\sqrt3 \) m
(b) 150 m
(c) 300 m
(d) \( 100\sqrt3 \) m
Show Answer & Explanation

Answer: (b) 150 m

Explanation:
1. Horizontal distance = 300 cos 60° = 150 m.

Teacher's Note:
1. Use cos for the horizontal component.

Question 12: A tower stands vertically on the ground. From a point 20 m from its foot the angle of elevation of the top is 60°. The height of the tower is:
(a) \( 20\sqrt3 \) m
(b) \( \frac{20}{\sqrt3} \) m
(c) 20 m
(d) 40 m
Show Answer & Explanation

Answer: (a) \( 20\sqrt3 \) m

Explanation:
1. Height = 20 tan 60° = 20√3 m.

Teacher's Note:
1. Use tan.

Question 13: The angle of elevation of the top of a tower at a distance of 150 m from its foot on level ground is 60°. The height of the tower is:
(a) \( 50\sqrt3 \) m
(b) 150 m
(c) \( 75\sqrt3 \) m
(d) \( 150\sqrt3 \) m (about 259.8 m)
Show Answer & Explanation

Answer: (d) \( 150\sqrt3 \) m (about 259.8 m)

Explanation:
1. Height = 150 tan 60° = 150√3 ≈ 259.8 m.

Teacher's Note:
1. Use tan.

Question 14: A tree is broken by the wind so that its top touches the ground at an angle of 30° with the ground, at a distance of 30 m from the root. The whole height of the tree before it broke is:
(a) 57.96 m
(b) 34.64 m
(c) \( 30\sqrt3 \) m (about 51.96 m)
(d) 60 m
Show Answer & Explanation

Answer: (c) \( 30\sqrt3 \) m (about 51.96 m)

Explanation:
1. Standing part = 30 tan 30° = 10√3 m. Broken part (hypotenuse) = 30/cos 30° = 20√3 m. Total = 30√3 ≈ 51.96 m.

Teacher's Note:
1. The tree height is the standing part plus the broken part.
2. The broken part is the slant side.

Question 15: From an aeroplane vertically above a straight horizontal road, the angles of depression of two stones 1 km apart on opposite sides of the aeroplane are α and β. The height of the aeroplane above the road (in km) is:
(a) \( \frac{\tan\alpha+\tan\beta}{\tan\alpha\tan\beta} \)
(b) \( \tan\alpha+\tan\beta \)
(c) \( \frac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta} \)
(d) \( \frac{\tan\alpha}{\tan\alpha+\tan\beta} \)
Show Answer & Explanation

Answer: (c) \( \frac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta} \)

Explanation:
1. The foot of the vertical is at distances h cot α and h cot β from the two stones. These add to 1: h(cot α + cot β) = 1, so h = 1/(cot α + cot β) = tan α tan β/(tan α + tan β).

Teacher's Note:
1. Write the two horizontal distances in terms of h.
2. They sum to 1 km.

Question 16: The angles of elevation of the top of a tower from two points at distances a and b from its base, on the same straight line through the base and on the same side, are complementary. The height of the tower is:
(a) \( ab \)
(b) \( \sqrt{ab} \)
(c) \( \frac{a+b}{2} \)
(d) \( \sqrt{a^2+b^2} \)
Show Answer & Explanation

Answer: (b) \( \sqrt{ab} \)

Explanation:
1. tan θ = h/a and tan(90° − θ) = h/b. Their product is 1: h²/(ab) = 1, so h = √(ab).

Teacher's Note:
1. tan θ × cot θ = 1.

Question 17: A vertical tower stands on level ground and carries a vertical flagstaff of height h on its top. From a point on the ground the angles of elevation of the bottom and the top of the flagstaff are α and β. The height of the tower is:
(a) \( \frac{h\tan\alpha}{\tan\beta-\tan\alpha} \)
(b) \( \frac{h\tan\beta}{\tan\beta-\tan\alpha} \)
(c) \( \frac{h\tan\alpha}{\tan\beta+\tan\alpha} \)
(d) \( h(\tan\beta-\tan\alpha) \)
Show Answer & Explanation

Answer: (a) \( \frac{h\tan\alpha}{\tan\beta-\tan\alpha} \)

Explanation:
1. Let the tower be H and the distance x. H = x tan α and H + h = x tan β. Subtracting, h = x(tan β − tan α), so x = h/(tan β − tan α) and H = h tan α/(tan β − tan α).

Teacher's Note:
1. Use the same distance x in both triangles.

Question 18: From a building 60 m high, the angles of depression of the top and the bottom of a lamp post are 30° and 60°. The distance between the lamp post and the building is:
(a) \( 60\sqrt3 \) m
(b) 20 m
(c) \( \frac{60}{\sqrt3}+20 \) m
(d) \( 20\sqrt3 \) m
Show Answer & Explanation

Answer: (d) \( 20\sqrt3 \) m

Explanation:
1. Distance = 60/tan 60° = 20√3 m (about 34.64 m).

Teacher's Note:
1. The bottom of the post gives the horizontal distance.

Question 19: From the top of a lighthouse 100 m above sea level, the angle of depression of a ship sailing towards it changes from 30° to 60°. The distance travelled by the ship is about:
(a) 173.2 m
(b) 115.47 m
(c) 57.74 m
(d) 100 m
Show Answer & Explanation

Answer: (b) 115.47 m

Explanation:
1. Distances: 100/tan 30° = 173.21 m and 100/tan 60° = 57.74 m. Difference = 115.47 m.

Teacher's Note:
1. Subtract the two horizontal distances.

Question 20: A tower is 50 m high. Its shadow is x m shorter when the sun's altitude is 45° than when it is 30°. The value of x, to the nearest hundredth, is:
(a) 36.60
(b) 86.60
(c) 50.00
(d) 28.87
Show Answer & Explanation

Answer: (a) 36.60

Explanation:
1. Shadows: 50/tan 30° = 86.60 m and 50/tan 45° = 50 m. x = 86.60 − 50 = 36.60 m.

Teacher's Note:
1. Find the shadow at each altitude.

Question 21: The angles of depression of the top and the bottom of a tower, seen from the top of a cliff 100 m high, are 30° and 60°. The height of the tower is about:
(a) 33.33 m
(b) 57.74 m
(c) 100 m
(d) 66.67 m
Show Answer & Explanation

Answer: (d) 66.67 m

Explanation:
1. Distance = 100/tan 60° = 100/√3 m. Drop to the tower top = (100/√3) tan 30° = 100/3 m. Height = 100 − 100/3 = 66.67 m.

Teacher's Note:
1. The bottom's angle fixes the distance.

Question 22: The angle of elevation of a cloud from a point 200 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°. The height of the cloud above the lake is:
(a) 200 m
(b) \( 200\sqrt3 \) m
(c) 400 m
(d) 600 m
Show Answer & Explanation

Answer: (c) 400 m

Explanation:
1. Let the cloud be h above the lake, horizontal distance x. Then h − 200 = x tan 30° and h + 200 = x tan 60°. Dividing: (h + 200)/(h − 200) = 3, so h = 400 m.

Teacher's Note:
1. The reflection is as far below the lake surface as the cloud is above it.
2. Divide the two equations.

Question 23: The angle of elevation of an aeroplane from a point P on the ground is 60°. After 15 seconds of level flight at a height of \( 1500\sqrt3 \) m, the elevation is 30°. The speed of the aeroplane is:
(a) 360 km/h
(b) 720 km/h
(c) 200 km/h
(d) 1200 km/h
Show Answer & Explanation

Answer: (b) 720 km/h

Explanation:
1. Horizontal distances: 1500√3/tan 60° = 1500 m and 1500√3/tan 30° = 4500 m. Distance flown = 3000 m in 15 s = 200 m/s = 720 km/h.

Teacher's Note:
1. Convert m/s to km/h by multiplying by 3.6.

Question 24: A bridge across a river makes an angle of 45° with the river bank. The length of the bridge is 150 m. The width of the river is:
(a) \( 75\sqrt2 \) m
(b) 150 m
(c) \( 150\sqrt3 \) m
(d) 75 m
Show Answer & Explanation

Answer: (a) \( 75\sqrt2 \) m

Explanation:
1. Width = 150 sin 45° = 150/√2 = 75√2 m.

Teacher's Note:
1. The width is perpendicular to the bank.

Free study material for Mathematics

Chapter 09 Some Applications of Trigonometry Objective Questions & Solutions for Class 10 Mathematics

Class 10 Mathematics Chapter 09 Some Applications of Trigonometry Objective Test Questions

Explore reliable practice questions for Chapter 09 Some Applications of Trigonometry tailored for Class 10 Mathematics learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

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