Trigonometry CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Trigonometry CBQs

1. Question 1MCQ1 mark
   A school builds a straight wheelchair ramp from the ground to a stage. The stage is 1 m higher than the ground and the sloping surface of the ramp is 2 m long. The angle the ramp makes with the horizontal ground is
   • 30°
   • 45°
   • 60°
   • 90°
   Show step-wise solution

   1. [1 mark] sin θ = rise ÷ sloping length = 1/2, and sin 30° = 1/2, so the ramp makes 30° with the ground.

   Answer: 30°

2. Question 2Assertion-Reason1 mark
   A straight ramp of sloping length L makes an angle θ with the horizontal ground, where 0° < θ < 90°. Assertion (A): The vertical rise of the ramp is always less than its sloping length L. Reason (R): sin θ < 1 for 0° < θ < 90°.
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (A): Rise = L sin θ and sin θ < 1 for an acute angle, so rise < L. Both are true and R explains A.

   Answer: (A)

3. Question 3Short answer2 marks
   A drone is hovering vertically above the main gate of a school during the annual sports day. Kavya stands on level ground 40 m from the gate and observes the drone at an angle of elevation of 60°. The school's safety rule says a drone must not fly higher than 60 m. Find the height of the drone and decide whether the rule is being followed. (Use √3 = 1.732)
   Show step-wise solution

   1. [1 mark] Let h be the height of the drone. tan 60° = h/40 ⇒ h = 40√3 m.
   2. [1 mark] h = 40 × 1.732 = 69.28 m, which is more than 60 m, so the drone is flying above the allowed height and the rule is not being followed.

   Answer: 40√3 m ≈ 69.28 m; rule not followed

4. Question 4Case study4 marks
   During the kite festival on Makar Sankranti, Arjun flies a kite from the flat terrace of his house. The kite string is 80 m long, taut and straight, and it makes an angle of 60° with the horizontal at Arjun's hand. Treat Arjun's hand as being at terrace level, and use √3 = 1.732 where needed. Based on the above information, answer the following questions: (i) Find the height of the kite above the terrace level. (ii) Find the horizontal distance between Arjun's hand and the point at terrace level directly below the kite. (iii) Later the wind weakens and, with the same 80 m of string still taut and straight, the string makes an angle of 30° with the horizontal. By how many metres has the height of the kite decreased?
   Show step-wise solution

   1. [1 mark] (i) sin 60° = h/80, so h = 80 × √3/2 = 40√3 m ≈ 69.28 m.
   2. [1 mark] (ii) cos 60° = x/80, so x = 80 × 1/2 = 40 m.
   3. [1 mark] (iii) New height = 80 sin 30° = 80 × 1/2 = 40 m.
   4. [1 mark] (iii) Decrease = 40√3 − 40 = 40(√3 − 1) = 40 × 0.732 = 29.28 m.

   Answer: (i) 40√3 m ≈ 69.28 m (ii) 40 m (iii) 40(√3 − 1) m ≈ 29.28 m

5. Question 5Long answer5 marks
   A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the tower? Give your answer to nearest minutes. (Use √3 = 1.73). OR If the angle of elevation of a cloud from a point 10 metres above a lake is 30° and the angle of depression of its reflection in the lake is 60°, find the height of the cloud from the surface of the lake.
   Show step-wise solution

   1. [1 mark] Let AB be the tower of height h and let the car move from D (angle of depression 30°) to C (angle of depression 45°) in 12 minutes; let the uniform speed give distance DC covered in 12 minutes.
   2. [1 mark] In ΔABC (depression 45°): tan 45° = AB/BC ⟹ BC = AB.
   3. [1 mark] In ΔABD (depression 30°): tan 30° = AB/BD ⟹ BD = √3·AB.
   4. [1 mark] DC = BD − BC = (√3 − 1)·AB, and this is covered in 12 minutes; time to cover BC = AB at the same speed = 12 × BC/DC = 12/(√3 − 1).
   5. [1 mark] Time = 12/(√3 − 1) = 6(√3 + 1) = 6(2.73) ≈ 16 minutes. [OR variant — cloud over a lake: observer 10 m above lake, horizontal distance x. tan 30° = (h − 10)/x and tan 60° = (h + 10)/x; dividing, 3 = (h + 10)/(h − 10) ⟹ 3h − 30 = h + 10 ⟹ h = 20 m.]

   Answer: ~16 minutes [or] cloud height = 20 m.

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