Class 10 Maths Chapter 8 Introduction to Trigonometry MCQs

Class 10 Maths MCQs for Chapter 8 (Introduction to trigonometry) are given here with answers and detailed explanations. All these multiple-choice questions are available online as per the CBSE syllabus and NCERT guidelines. Solving these objective questions will help students to score better marks in the board exam.

Class 10 Maths MCQs for Introduction to Trigonometry

Practice the MCQs for Chapter 8, introduction to the trigonometry of Class 10 Maths and verify your answers. Also, find important questions for class 10 Maths here to practice more.

Click here to download the PDF of additional MCQs for Practice on Introduction to Trigonometry Chapter of Class 10 Maths along with answer key:

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Students can also get access to Introduction to Trigonometry Class 10 Notes here.

1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of tan C is:

(a) 12/7

(b) 24/7

(c) 20/7

(d) 7/24

Answer: (b) 24/7

Explanation: AB = 24 cm and BC = 7 cm

tan C = Opposite side/Adjacent side

tan C = 24/7

2. (Sin 30°+cos 60°)-(sin 60° + cos 30°) is equal to:

(a) 0

(b) 1+2√3

(c) 1-√3

(d) 1+√3

Answer: (c) 1-√3

Explanation: sin 30° = ½, sin 60° = √3/2, cos 30° = √3/2 and cos 60° = ½

Putting these values, we get:

(½+½)-(√3/2+√3/2)

= 1 – [(2√3)/2]

= 1 – √3

3. The value of tan 60°/cot 30° is equal to:

(a) 0

(b) 1

(c) 2

(d) 3

Answer: (b) 1

Explanation: tan 60° = √3 and cot 30° = √3

Hence, tan 60°/cot 30° = √3/√3 = 1

4. 1 – cos2A is equal to:

(a) sin2A

(b) tan2A

(c) 1 – sin2A

(d) sec2A

Answer: (a) sin2A

Explanation: We know, by trigonometry identities,

sin2A + cos2A = 1

1 – cos2A = sin2A

5. sin (90° – A) and cos A are:

(a) Different

(b) Same

(c) Not related

(d) None of the above

Answer: (b) Same

Explanation: By trigonometry identities.

Sin (90°-A) = cos A {since 90°-A comes in the first quadrant of unit circle}

6. If cos X = ⅔ then tan X is equal to:

(a) 5/2

(b) √(5/2)

(c) √5/2

(d) 2/√5

Answer: (c) √5/2

Explanation: By trigonometry identities, we know:

1 + tan2X = sec2X

And sec X = 1/cos X = 1/(⅔) = 3/2

Hence,

1 + tan2X = (3/2)= 9/4

tan2X = (9/4) – 1 = 5/4

tan X = √5/2

7. If cos X = a/b, then sin X is equal to:

(a) (b2-a2)/b

(b) (b-a)/b

(c) √(b2-a2)/b

(d) √(b-a)/b

Answer: (c) √(b2-a2)/b

Explanation: cos X = a/b

By trigonometry identities, we know that:

sin2X + cos2X = 1

sin2X = 1 – cos2X = 1-(a/b)2

sin X = √(b2-a2)/b

8. The value of sin 60° cos 30° + sin 30° cos 60° is:

(a) 0

(b) 1

(c) 2

(d) 4

Answer: (b) 1

Explanation: sin 60° = √3/2, sin 30° = ½, cos 60° = ½ and cos 30° = √3/2

Therefore,

(√3/2) x (√3/2) + (½) x (½)

= (3/4) + (1/4)

= 4/4

= 1

9. 2 tan 30°/(1 + tan230°) =

(a) sin 60°

(b) cos 60°

(c) tan 60°

(d) sin 30°

Answer: (a) sin 60°

Explanation: tan 30° = 1/√3

Putting this value we get;

[2(1/√3)]/[1 + (1/√3)2] = (2/√3)/(4/3) = 6/4√3 = √3/2 = sin 60°

10. sin 2A = 2 sin A is true when A =

(a) 30°

(b) 45°

(c) 0°

(d) 60°

Answer: (c) 0°

Explanation: sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 0

11. The value of (sin 45° + cos 45°) is

(a) 1/√2

(b) √2

(c) √3/2

(d) 1

Answer: (b) √2

Explanation:

sin 45° + cos 45° = (1/√2) + (1/√2)

= (1 + 1)/√2

= 2/√2

= (√2 . √2)/√2

= √2

12. If sin A = 1/2 , then the value of cot A is

(a) √3

(b) 1/√3

(c) √3/2

(d) 1

Answer: (a) √3

Explanation:

Given,

sin A = 1/2

cos2A = 1 – sin2A

= 1 – (1/2)2

= 1 – (1/4)

= (4 – 1)/4

= 3/4

cos A = √(3/4) = √3/2

cot A = cos A/sin A = (√3/2)/(1/2) = √3

13. If ∆ABC is right angled at C, then the value of cos(A+B) is

(a) 0 

(b) 1

(c) 1/2

(d) √3/2

Answer: (a) 0

Explanation:

Given that in a right triangle ABC, ∠C = 90°.

We know that the sum of the three angles is equal to 180°.

∠A + ∠B + ∠C = 180° 

∠A + ∠B + 90° = 180° (∵ ∠C = 90° ) 

∠A + ∠B = 90° 

Now, cos(A+B) = cos 90° = 0

14. The value of (tan 1° tan 2° tan 3° … tan 89°) is

(a) 0 

(b) 1 

(c) 2 

(d) 1/2

Answer: (b) 1

Explanation:

tan 1° tan 2° tan 3°…tan 89°

= [tan 1° tan 2°…tan 44°] tan 45° [tan (90° – 44°) tan (90° – 43°)…tan (90° – 1°)]

= [tan 1° tan 2°…tan 44°] [cot 44° cot 43°…cot 1°] × [tan 45°]

= [(tan 1°× cot 1°) (tan 2°× cot 2°)…(tan 44°× cot 44°)] × [tan 45°]

= 1 × 1 × 1 × 1 × …× 1     {since tan A × cot A = 1 and tan 45° = 1}

= 1   

15. The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is

(a) -1 

(b) 0 

(c) 1 

(d) 3/2

Answer: (b) 0

Explanation:

[cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)]

= cosec [90° – (15° – θ)] – sec (15° – θ) – tan (55° + θ) + cot [90° – (55° + θ)] 

We know that cosec (90° – θ) = sec θ and cot(90° – θ) = tan θ.

= sec (15° – θ) – sec (15° – θ) – tan (55° + θ) + tan (55° + θ) 

= 0

16. If cos(α + β) = 0, then sin(α – β) can be reduced to

(a) cos β 

(b) cos 2β 

(c) sin α 

(d) sin 2α

Answer: (b) cos 2β 

Explanation:

Given,

cos(α + β) = 0

cos(α + β) = cos 90°

⇒ α + β = 90°

⇒ α = 90° – β 

sin(α – β) = sin(90° – β – β)

= sin(90° – 2β) 

= cos 2β    {since sin(90° – A) = cos A}

17. If sin A + sin2A = 1, then the value of the expression (cos2A + cos4A) is

(a) 1 

(b) 1/2 

(c) 2 

(d) 3

Answer: (a) 1

Explanation:

Given,

sin A + sin2A = 1

sin A = 1 – sin2A

sin A = cos2A {since sin2θ + cos2θ = 1}

Squaring on both sides,

sin2A = (cos2A)2

1 – cos2A = cos4A

⇒ cos2A + cos4A = 1

18. If cos 9α = sinα and 9α < 90°, then the value of tan 5α is

(a) 1/√3

(b) √3

(c) 1

(d) 0

Answer: (c) 1

Explanation:

Given,

cos 9α = sin α and 9α < 90°

That means, 9α is an acute angle.

cos 9α = cos(90° – α)

9α = 90° – α

9α + α = 90°

10α = 90°

α = 9°

tan 5α = tan(5 × 9°) = tan 45° = 1

19. The value of the expression \(\left [ \frac{sin^2(22^\circ)+sin^2(68^\circ)}{cos^2(22^\circ)+cos^2(68^\circ)}+sin^2(63^\circ)+cos\ 63^\circ\ sin\ 27^\circ \right ]\) is

(a) 3

(b) 2

(c) 1

(d) 0

Answer: (b) 2

Explanation:

Class 10 Maths Chapter 8 Introduction to trigonometry 19 A

20. The value of the expression sin6θ + cos6θ + 3 sin2θ cos2θ is

(a) 0

(b) 3

(c) 2

(d) 1

Answer: (d) 1

Explanation:

We know that, sin2θ + cos2θ = 1

Taking cube on both sides,

(sin2θ + cos2θ)3 = 1

(sin2θ)3 + (cos2θ)3 + 3 sin2θ cos2θ (sin2θ + cos2θ) = 1

sin6θ + cos6θ + 3 sin2θ cos2θ = 1

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