LIWA

Some Applications of Trigonometry

CBSE Class 10 · Maths · Chapter 9

Past board exam questions

CBSE 2027 sample paper · 1 mark

A flying kite is tied to a point on the ground with the help of a string. The string makes an angle θ with the ground level such that tan θ = (12)/(5). If the length of the string is 52 m, then the height of the kite above the ground is:

  1. 40 m
  2. 45.5 m
  3. 48 m
  4. 50 m
Show answer
48 m

CBSE 2027 sample paper · 5 marks

To assist an ambulance, a drone is hovering over an accident site at a constant height on a highway. The angle of depression of the ambulance, approaching the accident site, observed from drone is 30°. Twelve minutes later, the angle of depression of the ambulance is found to be 60°. If the ambulance is moving at a constant speed then, find the total time taken by the ambulance to reach the site of accident.

Show answer
Let the drone be at A, at height AB = h m above the accident site B. Let the ambulance be at D when the angle of depression is 30° and at C twelve minutes later (angle of depression 60°). Let its speed be x m/min, so CD = 12x m. In right ∆ABC: tan 60° = (AB)/(BC) ⇒ BC = (h)/(√(3)). In right ∆ABD: tan 30° = (AB)/(BD) ⇒ BD = h√(3). BD = BC + CD ⇒ (h)/(√(3)) + 12x = h√(3) ⇒ (2h)/(√(3)) = 12x ⇒ h = 6√(3)x. ∴ BC = (6√(3)x)/(√(3)) = 6x; time from C to B = (6x)/(x) = 6 minutes. Total time taken = 12 + 6 = 18 minutes.

CBSE 2027 sample paper · 5 marks

Manish is standing on level ground and observes a kite flying 200 m away from him at an angle of elevation of 30°. Mahesh, standing on the roof of a 50 m high building on the opposite side of the kite, observes the same kite at an angle of elevation of 45°. Find the distance between the kite and Mahesh.

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Figure: Manish at A on the ground; kite at B with E the foot of the perpendicular from B to the ground; building CD (D on the roof, where Mahesh stands, CD = 50 m) on the other side of E; R is the point on BE level with D, so RE = DC = 50 m. AB = 200 m (distance of the kite from Manish), ∠BAE = 30°, ∠BDR = 45°. In right ∆AEB: (BE)/(200) = sin 30° = (1)/(2) ⇒ BE = 100 m. BR = BE – RE = 100 – 50 = 50 m. In right ∆BRD: (50)/(BD) = sin 45° = (1)/(√(2)) ⇒ BD = 50√(2) m. The distance between the kite and Mahesh is 50√(2) m.

CBSE 2027 sample paper · 5 marks

A statue standing on a pedestal, is 1.6 m tall. From a point on the ground, the angles of elevation of the top and bottom of the statue are 60° and 45° respectively. Find the height of the pedestal.

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Let the pedestal be BC = h m, the statue CD = 1.6 m, and A the point on the ground. In right ∆ABC: tan 45° = (BC)/(AB) ⇒ 1 = (h)/(AB) ⇒ AB = h ...(i) In right ∆ABD: tan 60° = (BD)/(AB) ⇒ √(3) = (h + 1.6)/(AB) ⇒ AB = (h + 1.6)/(√(3)) ...(ii) From (i) and (ii): h = (h + 1.6)/(√(3)) ⇒ (√(3) - 1)h = 1.6 h = (1.6)/(√(3) - 1) = (1.6(√(3) + 1))/(2) = 0.8(√(3) + 1) m. ∴ Height of the pedestal is 0.8(√(3) + 1) m (≈ 2.19 m).

CBSE 2027 sample paper · 5 marks

The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be 45° and 30°. If the ships are 200m apart and one ship is exactly behind the other then find the height of lighthouse.

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Figure: lighthouse AB of height h m with top A; ships at C and D on the same side, C nearer (angle of depression 45°), D farther (30°), CD = 200 m. In right ∆ABC: (h)/(BC) = tan 45° = 1 ⇒ BC = h. In right ∆ABD: (h)/(h + 200) = tan 30° = (1)/(√(3)) ⇒ √(3)h = h + 200 ⇒ h = (200)/(√(3) - 1) × (√(3) + 1)/(√(3) + 1) = 100(√(3) + 1). The height of the lighthouse is 100(√(3) + 1) m (≈ 273.2 m).

CBSE 2026 board paper · 1 mark

If a pole is 6 m high and it casts a shadow of length 2√(3) m on the ground, then the Sun's elevation is :

  1. 60°
  2. 45°
  3. 120°
  4. 90°
Show answer
60°

CBSE 2026 board paper · 1 mark

The string of a flying kite is tied to a point on the ground. The length of the string between the kite and the point on the ground is 80 m. The string makes an angle of 30° with the ground. The height of the kite above the ground is :

  1. 20√(3) m
  2. 40 m
  3. 40√(3) m
  4. 80√(3) m
Show answer
40 m

CBSE 2026 board paper · 1 mark

If the length of the shadow of a tower is √(3) times that of its height, then altitude of the Sun is :

  1. 45°
  2. 30°
  3. 60°
  4. 15°
Show answer
30°