Chapter 9: Some Applications of Trigonometry Class 10 Maths NCERT Solutions

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Contents

Class 10 Maths NCERT Solutions - Chapter wise

1

Chapter 1: Real Numbers

2

Chapter 2: Polynomials

3

Chapter 3 : Pair of Linear Equations in Two Variables

4

Chapter 4 : Quadratic Equations

5

Chapter 5 : Arithmetic Progressions

6

Chapter 6 : Triangles

7

Chapter 7 : Coordinate Geometry

8

Chapter 8: Introduction to Trigonometry

9

Chapter 9: Some Applications of Trigonometry

10

Chapter 10 : Circles

11

Chapter 11 : Areas Related to Circles

12

Chapter 12: Surface Areas and Volumes

13

Chapter 13: Statistics

14

Chapter 14 : Probability

Chapter Notes

Chapter 9: Some Applications of Trigonometry — Exercise 9.1

Welcome to the real-world application of all those sin, cos, and tan ratios! This chapter, often called “Heights and Distances,” is how we calculate the height of a mountain or the width of a river without actually crossing them.

Q1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30 degrees.

  • The Logic: Visualize a right-angled triangle. The rope is the Hypotenuse (20 m), the pole is the Opposite side (height), and the angle is 30°. Since we need the Opposite side and have the Hypotenuse, we use the sin ratio (SOH: Sin = Opposite/Hypotenuse).

  • The Steps:

    • Let the height of the pole be ‘h’.

    • Angle = 30°. Hypotenuse = 20 m.

    • sin 30 = Opposite / Hypotenuse

    • 1/2 = h / 20

    • h = 20 / 2 = 10

  • The Result: The height of the pole is 10 m.

Q2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30 degrees 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.

  • The Logic: This is a two-part mission! The total height of the tree is the Standing part (Opposite) + the Broken part (Hypotenuse). We have the distance from the base (Adjacent = 8 m).

  • The Steps:

    1. Find the Standing part (h1) using tan 30:

      • tan 30 = h1 / 8 => 1 / root 3 = h1 / 8 => h1 = 8 / root 3.

    2. Find the Broken part (h2) using cos 30:

      • cos 30 = 8 / h2 => root 3 / 2 = 8 / h2 => h2 = 16 / root 3.

    3. Total height = h1 + h2 = (8 / root 3) + (16 / root 3) = 24 / root 3.

    4. Rationalize: (24 / root 3) x (root 3 / root 3) = 24 root 3 / 3 = 8 root 3.

  • The Result: The height of the tree is 8 root 3 m.

Q3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m… inclined at an angle of 30 degrees… whereas for elder children, she wants to have a steep slide at a height of 3 m… inclined at an angle of 60 degrees. What should be the length of the slide in each case?

  • The Logic: We need to find the Hypotenuse (length of slide) in two different triangles. In both cases, we are given the height (Opposite side). So, we stick with the sin ratio.

  • The Steps:

    • Case 1 (Below 5 years): height = 1.5 m, angle = 30°.

      • sin 30 = 1.5 / L1 => 1/2 = 1.5 / L1 => L1 = 3 m.

    • Case 2 (Elder children): height = 3 m, angle = 60°.

      • sin 60 = 3 / L2 => (root 3 / 2) = 3 / L2 => L2 = 6 / root 3.

      • Rationalize: L2 = 2 root 3 m.

  • The Result: The length of the slides are 3 m and 2 root 3 m.

Q4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30 degrees. Find the height of the tower.

  • The Logic: “Angle of Elevation” means you are looking up. We have the distance from the tower (Adjacent = 30 m) and need the height (Opposite). We use tan 30.

  • The Steps:

    • Let height = h.

    • tan 30 = h / 30

    • 1 / root 3 = h / 30

    • h = 30 / root 3 = 10 root 3 m.

  • The Result: The height of the tower is 10 root 3 m.

Q5. A kite is flying at a height of 60 m above the ground… the inclination of the string with the ground is 60 degrees. Find the length of the string, assuming that there is no slack in the string.

  • The Logic: The height of the kite is the Opposite side (60 m). The string is the Hypotenuse. Angle is 60°. We use sin 60.

  • The Steps:

    • Let length of string = L.

    • sin 60 = 60 / L

    • root 3 / 2 = 60 / L

    • L = 120 / root 3 = 40 root 3 m.

  • The Result: The length of the string is 40 root 3 m.

Q6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30 degrees to 60 degrees as he walks towards the building. Find the distance he walked towards the building.

  • The Logic: Since the boy is 1.5 m tall, his “eye level” is 1.5 m above the ground. The effective height of the building we use for our triangles is 30 – 1.5 = 28.5 m. We calculate the horizontal distance at 30° and at 60°, then find the difference.

  • The Steps:

    1. Effective height (h) = 28.5 m.

    2. At 30°: tan 30 = 28.5 / x1 => 1 / root 3 = 28.5 / x1 => x1 = 28.5 root 3.

    3. At 60°: tan 60 = 28.5 / x2 => root 3 = 28.5 / x2 => x2 = 28.5 / root 3 = 9.5 root 3.

    4. Distance walked = x1 – x2 = 28.5 root 3 – 9.5 root 3 = 19 root 3.

  • The Result: The distance walked towards the building is 19 root 3 m.

Q7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45 degrees and 60 degrees respectively. Find the height of the tower.

  • The Logic: We have a building (20 m) and a tower on top of it.

    1. Use the 45° angle to find the distance from the point to the building.

    2. Use the 60° angle and that distance to find the total height (Building + Tower).

  • The Steps:

    1. Distance to building (d): tan 45 = 20 / d => 1 = 20 / d => d = 20 m.

    2. Total height (H): tan 60 = H / 20 => root 3 = H / 20 => H = 20 root 3.

    3. Height of tower = Total height – Building height = 20 root 3 – 20.

  • The Result: The height of the tower is 20(root 3 – 1) m.

Q8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60 degrees and from the same point the angle of elevation of the top of the pedestal is 45 degrees. Find the height of the pedestal.

  • The Logic: This is the reverse of Q7! We know the statue (top part) and need the pedestal (bottom part).

  • The Steps:

    1. Let pedestal height = p. Distance from point = d.

    2. tan 45 = p / d => 1 = p / d => p = d.

    3. tan 60 = (p + 1.6) / d. Since p = d, we get: root 3 = (p + 1.6) / p.

    4. p root 3 = p + 1.6 => p(root 3 – 1) = 1.6.

    5. p = 1.6 / (root 3 – 1). Rationalize: p = 0.8(root 3 + 1).

  • The Result: The height of the pedestal is 0.8(root 3 + 1) m.

Q9. The angle of elevation of the top of a building from the foot of the tower is 30 degrees and the angle of elevation of the top of the tower from the foot of the building is 60 degrees. If the tower is 50 m high, find the height of the building.

  • The Logic: We have two objects standing opposite each other. The common link is the ground distance (d) between them.

     

  • The Steps:

    1. From Tower (50 m) at 60°: tan 60 = 50 / d => root 3 = 50 / d => d = 50 / root 3.

    2. From Building (h) at 30°: tan 30 = h / d => 1 / root 3 = h / (50 / root 3).

    3. h = (1 / root 3) x (50 / root 3) = 50 / 3.

  • The Result: The height of the building is 16 2/3 m (or 16.67 m).

Q10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60 degrees and 30 degrees, respectively. Find the height of the poles and the distances of the point from the poles.

  • The Logic: Let height of poles = h. Let the point be ‘x’ distance from the first pole and (80 – x) from the second.

  • The Steps:

    1. Pole 1 (60°): tan 60 = h / x => h = x root 3.

    2. Pole 2 (30°): tan 30 = h / (80 – x) => 1 / root 3 = h / (80 – x).

    3. Substitute h: 1 / root 3 = (x root 3) / (80 – x).

    4. 80 – x = 3x => 4x = 80 => x = 20 m.

    5. Height h = 20 root 3.

  • The Result: The height of the poles is 20 root 3 m and the point is 20 m and 60 m from the poles.

Q11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60 degrees. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30 degrees. Find the height of the tower and the width of the canal.

  • The Logic: We have two triangles sharing the same tower height (h). The width of the canal is x. When the observer moves 20 m further back, the base becomes x + 20.

  • The Steps:

    1. In the 60° triangle: tan 60 = h / x => root 3 = h / x => h = x root 3.

    2. In the 30° triangle: tan 30 = h / (x + 20) => 1 / root 3 = h / (x + 20).

    3. Substitute h: 1 / root 3 = (x root 3) / (x + 20).

    4. Cross multiply: x + 20 = 3x => 2x = 20 => x = 10 m.

    5. Height h = 10 root 3.

  • The Result: The height of the tower is 10 root 3 m and the width of the canal is 10 m.

Q12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60 degrees and the angle of depression of its foot is 45 degrees. Determine the height of the tower.

  • The Logic: You are standing on a building, looking up at the tower top and down at its base. The height of the building (7 m) gives you the first “slice” of the tower.

  • The Steps:

    1. Looking down (45°): The distance to the tower (d) is the same as the building height because tan 45 = 1. So, d = 7 m.

    2. Looking up (60°): Let the upper part of the tower be ‘h’.

      tan 60 = h / d => root 3 = h / 7 => h = 7 root 3.

    3. Total height = Building height + Upper part = 7 + 7 root 3.

  • The Result: The height of the tower is 7(root 3 + 1) m.

Q13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30 degrees and 45 degrees. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

  • The Logic: “Angle of Depression” is the same as the “Angle of Elevation” from the ship to the lighthouse (Alternate interior angles). We calculate the distance of both ships from the lighthouse and subtract them.

  • The Steps:

    1. Ship 1 (45°): tan 45 = 75 / x1 => 1 = 75 / x1 => x1 = 75 m.

    2. Ship 2 (30°): tan 30 = 75 / x2 => 1 / root 3 = 75 / x2 => x2 = 75 root 3.

    3. Distance between ships = x2 – x1 = 75 root 3 – 75.

  • The Result: The distance between the ships is 75(root 3 – 1) m.

Q14. A 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 any instant is 60 degrees. After some time, the angle of elevation reduces to 30 degrees. Find the distance travelled by the balloon during the interval.

  • The Logic: Like the boy in Q6, we subtract the girl’s height first: 88.2 – 1.2 = 87 m. The balloon is moving away, so the angle decreases from 60° to 30°.

  • The Steps:

    1. At 60°: tan 60 = 87 / x1 => root 3 = 87 / x1 => x1 = 87 / root 3 = 29 root 3.

    2. At 30°: tan 30 = 87 / x2 => 1 / root 3 = 87 / x2 => x2 = 87 root 3.

    3. Distance traveled = x2 – x1 = 87 root 3 – 29 root 3 = 58 root 3.

  • The Result: The distance travelled by the balloon is 58 root 3 m.

Q15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30 degrees, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60 degrees. Find the time taken by the car to reach the foot of the tower from this point.

  • The Logic: This is a “Time and Speed” puzzle. Instead of distances, we work with parts of the base.

     

  • The Steps:

    1. Let height = h. At 60°: Distance (d1) = h / root 3.

    2. At 30°: Total distance (d2) = h root 3.

    3. Distance covered in 6 seconds = d2 – d1 = h root 3 – h / root 3 = 2h / root 3.

    4. Distance remaining = d1 = h / root 3.

    5. Observe the ratio: The remaining distance is exactly half of the distance covered in 6 seconds.

  • The Result: Since the remaining distance is half, it will take half the time, which is 3 seconds.

Student Summary Table (The Final Masterlist)

Question Type The “Topper” Shortcut What to double-check?
Depression = Elevation Use Z-shape angles. Label angles correctly from top to bottom.
Moving Object Subtract distances. Is the object moving toward or away?
Speed/Time Use ratios of distances. Ensure the speed is uniform.
Height Slice Add/Subtract eye level. Check if observer height is given.

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