Class 10 Maths Chapter 10 Some Applications of Trigonometry Most Important Questions By PW

Let AC be the building & DE be the towerPhysics Wallah brings you a well-designed collection of Most Important Questions for Class 10 Maths Chapter 9: Some Applications of Trigonometry to strengthen your exam preparation. This chapter focuses on applying trigonometric ratios to real-life situations, like finding heights and distances using angles of elevation and depression. 

With PW’s targeted questions, you learn to approach problems logically and avoid common calculation errors. Practising these carefully selected questions not only improves your conceptual clarity but also enhances your speed and accuracy, helping you handle exam-level questions with greater confidence and precision.

CBSE Class 10 Sample Paper 

Important Questions for Class 10 Maths Chapter 10 Some Applications of Trigonometry

Q.1: 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 between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution:

Using given instructions, draw a figure. Let AC be the broken part of the tree. Angel C = 30 degrees. BC = 8 m To Find: Height of the tree, which is AB

Q.2: A tower stands vertically on the ground. From a point on the ground which is 25 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 45°. Then the height (in meters) of the tower is :

(a) 25 2 2 ​ (b) 25 3 3 ​ (c) 25 (d) 12.5

Solution:

Let the height of the tower be H m.

Thus, tan45°= Base/ perpendicular Or H/25 =1 Hence, H = 25m

Q.3: A ladder, leaning against a wall, makes anangle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder. (2016OD)

Solution:

Let AC be the ladder

∴ Length of ladder, AC = 5 m 2.5 m

Q.4: A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then calculate the height of the wall.

Solution:

∠BAC = 180° – 90° – 60o = 30° sin 30° = BCAC 12=BC15 2BC = 15 BC = 152m

Q.4: In the given figure, a tower AB is 20 m high and BC, its shadow on the ground, is 203–√ m long. Find the Sun’s altitude.

Solution:

AB = 20 m, BC = 20 3 –√ m,

θ = ? In ∆ABC

Q.5: The angles of depression of two ships from the top of a light house and on the same side of it are found to be 45° and 30°. If the ships are 200 m apart, find the height of the light house.

Solution:

Let AB be the height of the light house, D and C are two ships and DC = 200 m Let BC = x m, AB = h m In rt. ∆ABC,∴ Height of the light house = 273 m

Q.6: The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will

(A) also get doubled (B) will get halved (C) will be less than 60 degree (D) None of these

Solution:

According to Question:

Q.7: If the height of a tower and the distance of the point of observation from its foot,both, are increased by 10%, then the angle of elevation of its top

(A) increases (B) decreases (C) remains unchanged (D) have no relation.

Solution:

Since tan θ = h/x Where h is height and x is distance from tower, If both are increased by 10%, then the angle will remain unchanged.

Q.8: As observed from the top of a 60 m high light house 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 light-house, find the distance between the two ships. (Use 3 –√ = 1.732]

Solution:

Let AB = 60 m be the height of Light-house and C and D be the two ships. In right ∆ABC,

∴ Distance between the two ships, CD = BD – BC = 103.92 – 60 = 43.92 m

Q.9: The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 3000 3 –√ m, find the speed of the aeroplane.

Solution:

Let A be the point on the ground and C be the aeroplane. In rt. ∆ABC

Q.10: From the top of a 60 m high building, the angles of depression of the top and the bottom of a tower are 45° and 60° respectively. Find the height of the tower. (Take 3 –√ = 1.73] (2014OD)

Solution:

Let AC be the building & DE be the tower.

.

∴ Height of the tower, DE = BC DE = AC – AB DE = 60 – 20 3 –√ = 20(3 – 3 –√ ) DE = 20(3 – 1.73) = 20(1.27) DE = 25.4 m

Q.11. The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building. [Use 3 –√ = 1.73].

Solution:

Let AE be the building and CD be the tower. Let height of the tower = h m and, the horizontal distance between tower and building = x m …[Given BD = AE = 50 m ∴ BC = CD – BD = (h – 50) m

From (i), x = h – 50 = 118.25 – 50 = 68.25 m Height of the tower, h = 118.25 m ∴ Horizontal distance between tower and Building, x = 68.25 m

Q.12: A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of hill as 30°. Find the distance of the hill from the ship and the height of the hill. (2016D)

Solution:

Let the man standing on the deck of a ship be at point A and let CE be the hill. Here BC is the distance of hill from ship and CE be the height of hill. In rt. ∠ABC, tan 30° = A B B C BC = 10 3 –√ m .(i) BC = 10(1.73) = 17.3 m …[:: 3 –√ = 1.73 AD = BC = 10 3 –√ m …(ii) [From (i) In rt. ∆ADE, tan 60° = D E A D ⇒ 3 –√ = D E 10 3 √ … [From (ii) ⇒ DE = 10 3 –√ × 3 –√ = 30 m ∴ CE = CD + DE = 10 + 30 = 40 m Hence, the distance of the hill from the ship is 17.3 m and the height of the hill is 40 m.

Q.13: From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars. (Use 3 –√ = 1.732]

Solution:

Let AB be the tower. In rt. ∆ABC, tan 45° = A B B C

∴ Distance between the cars, CD = BD + BC = 173 + 100 = 273 m

Q.14: Two poles of equal height are standing opposite to each other on either side of the road, which is 100 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles.

Solution:

Let AB and DE be the two equal poles and C be the point on BD (road). Let BC = x m Then CD = (100 – x) m Let AB = DE = y m In rt. ∆ABC,⇒ 3y = 100 3 –√ – y ⇒ 4y = 100 3 –√ ∴ Height of the poles, y = 100 3 √ 4 = 25 3 –√ m = 25(1.73) = 43.25 m

Q.15: 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, find the height of the building.

Solution:

Let AB = 50 m be the tower and CD be the building. In rt. ∆ABC,