LIWA

Introduction to Trigonometry

CBSE Class 10 · Maths · Chapter 8

Past board exam questions

CBSE 2027 sample paper · 1 mark

If tan (A + B) = √(3) and tan (A – B) = (1)/(√(3)), 0° < A + B < 90°, A > B, then the value of A and B respectively are:

  1. 60°, 30°
  2. 60°, 45°
  3. 45°, 15°
  4. 60°,15°
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45°, 15°

CBSE 2027 sample paper · 1 mark

The value of sin2 60° - 2tan2 45° - cos2 30° is

  1. −2
  2. −1
  3. 1
  4. 2
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−2

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
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48 m

CBSE 2027 sample paper · 2 marks

If 3sinθ = 4cosθ, find the value of sinθ + cos2θ - 1; 0° < θ < 90°

Show answer
(sinθ)/(cosθ) = (4)/(3) ⇒ tanθ = (4)/(3), so sinθ = (4)/(5) and cosθ = (3)/(5). ∴ sinθ + cos2θ - 1 = (4)/(5) + (9)/(25) - 1 = (20 + 9 - 25)/(25) = (4)/(25).

CBSE 2027 sample paper · 2 marks

If tanθ = (1)/(√(5)), then find the value of (cosec2θ - sec2θ)/(cosec2θ + sec2θ).

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tanθ = (1)/(√(5)) ⇒ cotθ = √(5). sec2θ = 1 + tan2θ = 1 + (1)/(5) = (6)/(5) and cosec2θ = 1 + cot2θ = 1 + 5 = 6. (cosec2θ - sec2θ)/(cosec2θ + sec2θ) = (6 - (6)/(5))/(6 + (6)/(5)) = (24/5)/(36/5) = (2)/(3).

CBSE 2027 sample paper · 3 marks

Consider ∆ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ABC = θ. Determine the values of i. 1 + tan2θ ii. cos2θ - sin2θ

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AC = √(AB2 - BC2) = √(292 - 212) = √(400) = 20 units. sinθ = (AC)/(AB) = (20)/(29), cosθ = (BC)/(AB) = (21)/(29), tanθ = (AC)/(BC) = (20)/(21). (i) 1 + tan2θ = 1 + ((20)/(21))2 = (441 + 400)/(441) = (841)/(441) (ii) cos2θ - sin2θ = ((21)/(29))2 - ((20)/(29))2 = (441 - 400)/(841) = (41)/(841)

CBSE 2027 sample paper · 3 marks

During a math class, Ms. Isha wrote the expression given below on the board and asked the students to simplify it. (cosθ)/(1 - sinθ) + (1 - sinθ)/(cosθ) Jyoti solved it in her note book as follows: (cosθ)/(1 - sinθ) + (1 - sinθ)/(cosθ) = (cos2θ + (1 - sinθ)2)/((1 - sinθ) × cosθ) …. (step 1) = (cos2θ + cos2θ)/((1 - sinθ) × cosθ) …. (step 2) = (2cos2θ)/((1 - sinθ) × cosθ) …. (step 3) = (2cosθ)/((1 - sinθ)) …. (step 4) Identify the step in which Jyoti has made error(s), if any. Rectify the same and find the correct answer.

Show answer
Jyoti made the error in step 2: (1 - sinθ)2 ≠ cos2θ. Correct solution: (cosθ)/(1 - sinθ) + (1 - sinθ)/(cosθ) = (cos2θ + (1 - sinθ)2)/((1 - sinθ)cosθ) = (cos2θ + 1 + sin2θ - 2sinθ)/((1 - sinθ)cosθ) = (2 - 2sinθ)/((1 - sinθ)cosθ) = (2(1 - sinθ))/((1 - sinθ)cosθ) = (2)/(cosθ) = 2secθ.

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.