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Some Applications of Trigonometry Class 10 Maths Worksheet with Answers Chapter 9

Updated:10 Mins Read

Some Applications of Trigonometry are in continuation from the last chapter, Introduction to Trigonometry. In this class 10 NCERT book based worksheet is divided into four steps: Basic for definitions and quick setups, Standard for direct height–distance calculations, Advance for multi-step setups, and HOTS for tricky reasoning questions that test your diagram skills. Draw first, solve next. Use the hints and answers only after your own attempt.

Topics covered in the worksheet

  • Angle of elevation
  • Angle of depression
  • Heights and distances
  • Right-triangle modelling
  • Real-life applications

Class 10 Maths Worksheet – Chapter 9: Some Applications of Trigonometry

Basic

  1. Define in one line:

    • Line of sight
    • Angle of elevation
    • Angle of depression
  2. A student looks at the top of a tower. The line of sight makes an angle above the horizontal.
    Is it elevation or depression?
  3. From the top of a building, you look down at a car on the road.
    Is it elevation or depression?
  4. A vertical pole is 12 m high. From a point on the ground 12 m away from its foot, the angle of elevation is 45°.
    Find the height of the pole.
  5. A tower is viewed from a point 10√3 m away. The angle of elevation of its top is 30°.
    Find the height of the tower.
  6. A kite is flying at a height of 30 m. The angle of elevation from the point where the string is held is 30°.
    Find the length of the string (assume it is tight).
  7. A ladder makes an angle of 60° with the ground and reaches a point 3 m high on a wall.
    Find the length of the ladder.

Standard

  1. From a point 25 m away from the foot of a tower, the angle of elevation of the top is 60°.
    Find the height of the tower.
  2. A 1.6 m tall person stands 20 m away from a building. The angle of elevation of the top of the building from the person’s eyes is 45°.
    Find the height of the building.
  3. A ladder is placed against a wall. It reaches a point 4.5 m high and makes an angle of 30° with the ground.
    Find:

    • Length of ladder
    • Distance of its foot from wall
  4. A flagstaff stands on a 12 m building. From a point P, the angle of elevation of the top of the building is 30°, and that of the top of the flagstaff is 45°.
    Find:

    • Distance of P from building
    • Length of flagstaff
  5. The shadow of a pole is 8 m when the Sun’s altitude is 45°.
    Find the height of the pole.
  6. From the top of a 10 m building, the angle of depression of a car on the ground is 45°.
    Find the distance of the car from the building.
  7. A cable is stretched from the top of a 15 m tower to a point on the ground. The angle of elevation from that point to the top is 60°.
    Find the length of the cable.

Advance

  1. The shadow of a tower is 30 m when the Sun’s altitude is 45°. Later, the shadow becomes 30 + 10√3 m when the Sun’s altitude is 30°.
    Find the height of the tower.
  2. From a point on the ground, the angle of elevation of the top of a 20 m building is 30°. From the same point, the angle of elevation of the top of a tower standing on the building is 60°.
    Find the height of the tower.
  3. From the top of a multi-storeyed building, the angles of depression of the top and the bottom of an 8 m building are 30° and 45° respectively.
    Find:

    • Height of the multi-storeyed building
    • Distance between the buildings
  4. From a point A on the ground, the angle of elevation of the top of a tower is 30°. From another point B, which is 20 m closer to the tower than A, the angle of elevation is 60°.
    Find the height of the tower.
  5. From the top of a 7 m building, the angle of elevation of the top of a cable tower is 60°, and the angle of depression of its foot is 45°.
    Find the height of the cable tower.
  6. A balloon is at a height of 60 m from the ground. The angle of elevation from point P is 60°. After some time, the angle becomes 30° (balloon moves horizontally).
    Find the horizontal distance travelled by the balloon.
  7. A tree breaks and the top touches the ground making an angle of 30° with the ground. The distance from the foot of the tree to the touching point is 12 m.
    Find the original height of the tree.

HOTS

  1. From a point on a bridge 4 m above the river banks, the angles of depression of the banks on opposite sides are 30° and 45°.
    Find the width of the river.
  2. Two poles of equal height stand on opposite sides of a road 60 m wide. From a point between them, the angles of elevation of the tops are 60° and 30° respectively.
    Find:

    • Height of each pole
    • Distances of the point from the poles
  3. A lighthouse is 50 m high. The angles of depression of two ships on the same side are 30° and 45° (one behind the other).
    Find the distance between the two ships.
  4. A car is approaching a tower. From the top of the tower, the angle of depression of the car changes from 30° to 60° in 6 seconds.
    If the car moves with uniform speed, find the time taken (after the 60° position) to reach the foot of the tower.
  5. A 1.5 m tall boy observes the top of a 30 m building. His angle of elevation changes from 30° to 60° as he walks towards the building.
    Find how far he walked.
  6. A statue (2 m tall) is on top of a pedestal. From a point on the ground, the angle of elevation of the top of the pedestal is 45°, and of the top of the statue is 60°.
    Find the height of the pedestal.
  7. A TV tower stands on one bank of a canal. From a point directly opposite on the other bank, angle of elevation is 60°. From another point 15 m away from this point (on the line towards the tower), angle of elevation is 30°.
    Find:

    • Height of the tower
    • Width of the canal

Answer Key

Basic – Answers

  1. Ans:

    • Line of sight: Line from observer’s eye to the object point.
    • Elevation: Angle with horizontal when looking upward.
    • Depression: Angle with horizontal when looking downward.

    Hint: Horizontal line is your reference.

  2. Ans: Angle of elevation.
    Hint: Line of sight is above horizontal.
  3. Ans: Angle of depression.
    Hint: Looking down means depression.
  4. Ans: Height = 12 m.
    Hint: tan 45° = h/12 ⇒ h = 12.
  5. Ans: Height = 10 m.
    Hint: tan 30° = h/(10√3) = 1/√3 ⇒ h = 10.
  6. Ans: String length = 60 m.
    Hint: sin 30° = 30/L ⇒ 1/2 = 30/L ⇒ L = 60.
  7. Ans: Ladder length = 2√3 m.
    Hint: sin 60° = 3/L ⇒ √3/2 = 3/L ⇒ L = 6/√3 = 2√3.

Standard – Answers

  1. Ans: Height = 25√3 m.
    Hint: tan 60° = h/25 ⇒ √3 = h/25.
  2. Ans: Height of building = 21.6 m.
    Hint: tan 45° = (H − 1.6)/20 ⇒ H = 20 + 1.6.
  3. Ans:

    • Ladder = 9 m
    • Foot from wall = 9√3/2 m

    Hint: sin 30° = 4.5/L ⇒ L = 9; then cos 30° gives base.

  4. Ans:

    • Distance P from building = 12√3 m
    • Flagstaff length = 12(√3 − 1) m

    Hint: Use tan 30° for building, tan 45° for total height.

  5. Ans: Height = 8 m.
    Hint: tan 45° = h/8 ⇒ h = 8.
  6. Ans: Distance = 10 m.
    Hint: tan 45° = 10/x ⇒ x = 10.
  7. Ans: Cable length = 10√3 m.
    Hint: sin 60° = 15/L ⇒ √3/2 = 15/L.

Advance – Answers

  1. Ans: Height = 30 m.
    Hint: From 45°: h = 30. Check with 30° shadow: h√3 = 30 + 10√3.
  2. Ans: Tower height = 20(√3 − 1) m.
    Hint: Find distance x from tan 30° with 20 m, then use tan 60° for (20 + tower).
  3. Ans:

    • Height = 8(1 + √3) m
    • Distance = 8(1 + √3) m

    Hint: Use tan 45° and tan 30° with alternate angles logic.

  4. Ans: Height = 10√3 m.
    Hint: Let distance from A be x and from B be (x − 20). Use tan 30° and tan 60°.
  5. Ans: Height of tower = 7 + 7√3 m.
    Hint: From depression 45°: horizontal distance = 7. Then use elevation 60° to get extra height.
  6. Ans: Distance travelled = 60√3 m.
    Hint: x₁ = 60/√3, x₂ = 60√3. Travel = x₂ − x₁ = 60(√3 − 1/√3) = 60·(2/√3) = 40√3? (Wait) Do carefully:
    x₁ = 60/ tan60° = 60/√3 = 20√3
    x₂ = 60/ tan30° = 60/(1/√3) = 60√3
    Travel = 60√3 − 20√3 = 40√3 m.
    Ans: 40√3 m.
  7. Ans: Original height = 12 + 12/√3 = 12 + 4√3 m.
    Hint: Broken part length = 12/ cos30° = 24/√3 = 8√3. Total = 12 + 8√3? Wait:
    Let stump = h, broken part = L, base distance = 12.
    Angle with ground = 30° ⇒ cos30° = 12/L ⇒ L = 12/(√3/2) = 24/√3 = 8√3.
    Height of stump = L·sin30° = 8√3·(1/2) = 4√3.
    Total height = h + L = 4√3 + 8√3 = 12√3 m.
    Ans: 12√3 m.

HOTS – Answers

  1. Ans: Width = 4(1 + √3) m.
    Hint: AD = 4√3 from tan30°, BD = 4 from tan45°, AB = AD + BD.
  2. Ans:

    • Height = 20√3 m
    • Distances = 20 m and 40 m

    Hint: Let distances be x and (60 − x). Use tan60° and tan30° and equate heights.

  3. Ans: Distance between ships = 50(√3 − 1) m.
    Hint: Distances from lighthouse base are 50/ tan45° = 50 and 50/ tan30° = 50√3.
  4. Ans: Time = 6 seconds.
    Hint: Let tower height = h.
    At 30°: distance = h/ tan30° = h√3
    At 60°: distance = h/ tan60° = h/√3
    In 6 s, car covers h√3 − h/√3 = 2h/√3
    Remaining to foot from 60° point = h/√3
    So remaining time = (h/√3) ÷ (2h/√3 / 6) = 3 s.
    Ans: 3 seconds.
  5. Ans: Distance walked = 15√3 m.
    Hint: Effective height = 30 − 1.5 = 28.5.
    d₁ = 28.5/ tan30° = 28.5√3, d₂ = 28.5/ tan60° = 28.5/√3.
    Walked = d₁ − d₂ = 28.5(√3 − 1/√3) = 28.5·(2/√3) = 19? Do carefully:
    d₁ − d₂ = 28.5√3 − 28.5/√3 = 28.5·( (3 − 1)/√3 ) = 28.5·(2/√3) = 57/√3 = 19√3 m.
    Ans: 19√3 m.
  6. Ans: Pedestal height = 2(1 + √3) m.
    Hint: Let pedestal = h, distance = x.
    tan45° = h/x ⇒ x = h.
    tan60° = (h + 2)/x ⇒ √3 = (h + 2)/h ⇒ h(√3 − 1) = 2.
    h = 2/(√3 − 1) = 2(√3 + 1)/2 = √3 + 1.
    Ans: h = (1 + √3) m. (Pedestal height)
  7. Ans:

    • Height = 5√3 m
    • Canal width = 5 m

    Hint: Let width = w.
    From opposite point: tan60° = h/w ⇒ h = w√3.
    From 15 m away point: distance = w + 15, tan30° = h/(w + 15) = 1/√3.
    Substitute h = w√3 ⇒ w√3 /(w + 15) = 1/√3 ⇒ 3w = w + 15 ⇒ w = 7.5? (Check)
    3w = w + 15 ⇒ 2w = 15 ⇒ w = 7.5, h = 7.5√3.
    Ans: Width = 7.5 m, Height = 7.5√3 m.

Worksheet for Other chapters

Amit, a BE in Mechanical Engineering, is a math enthusiast dedicated to making math fun and accessible for kids in classes 1 to 10. With a knack for simplifying complex concepts, Amit offers easy-to-understand solutions, fostering a love for math in young minds across India.

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