Class 10 Maths MCQ Question of Introduction to trigonometry with answers and detailed explanations. These MCQ Questions are available online as per the CBSE syllabus and NCERT guidelines. Solving these objective questions will help students to attain better marks within the board exam.
Below you’ll find Multiple Choice Questions of Introduction to Trigonometry Class 10 Maths which will assist you in gaining good marks within the examinations and also cracking competitive exams. These MCQ Questions for Class 10 Maths with answers will increase your skills and understand concepts in a better manner.
1. The study of relationships between the sides and angles of a triangle is….
(a) Statistics
(b) Trigonometry
(c) Geometry
(d) None of the above
2. The value of tan 45 is ——
(a) 1
(b) 0
(c) 1/2
(d)\(\sqrt 3\)
3. 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
4. The value of tan 60°/cot 30° is equal to:
(a) 0
(b) 1
(c) 2
(d) 3
5. 1-cos\(^2\)A is equal to:
(a) sin\(^2\)A
(b) tan\(^2\)A
(c) 1-sin\(^2\)A
(d) sec\(^2\)A
6. Sin (90° – A) and cos A are:
(a) Different
(b) Same
(c) Not related
(d) None of the above
7. If x and y are complementary angles, then
(a) sin x = sin y
(b) tan x = tan y
(c) cos x = cos y
(d) sec x = cosec y
8. What is the minimum value of sin A, 0 ≤ A ≤ 90°
(a) -1
(b) 0
(c) 1
(d) \(\frac{1}{2}\)
9. What is the minimum value of cos θ, 0 ≤ θ ≤ 90°
(a) -1
(b) 0
(c) 1
(d) \(\frac{1}{2}\)
10. Ratios of sides of a right triangle with respect to its acute angles are known as
(a) trigonometric identities
(b) trigonometry
(c) trigonometric ratios of the angles
(d) none of these
11. 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
12. If cos 9α = sin a and 9α < 90°, then the value of tan 5α is
(a) \(\frac{1}{\sqrt3}\)
(b) \(\sqrt3\)
(c) 1
(d) 0
13. sin (45° + θ) – cos (45° – θ) is equal to
(a) 2 cos θ
(b) 0
(c) 2 sin θ
(d) 1
14. sec\(^2\)θ – 1 = ?
(a) tan\(^2\)θ
(b) tan\(^2\)θ + 1
(c) cot\(^2\)θ – 1
(d) cos\(^2\)θ
15. The value of sin θ and cos (90° – θ)
(a) Are same
(b) Are different
(c) No relation
(d) Information insufficient
16. The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is
(a) 1
(b) −1
(c) 0
(d) 1/2
17. Out of the following options, the two angles that are together classified as complementary angles are
(a) 120° and 60°
(b) 50° and 30°
(c) 65° and 25°
(d) 70° and 30°
18. If sin θ − cos θ = 0, vthen the value of θ is
(a) 90°
(b) 30°
(c) 45°
(d) 60°
19. sin 2A = 2 sin A is true when A =
(a) 30°
(b) 45°
(c) 0°
(d) 60°
20. Reciprocal of tan A is —
(a) cosec A
(b) sec A
(c) cot A
(d) None of the above
Answers & Explanations
1. Answer: (b) Trigonometry
2. Answer: (a) 1
3. Answer: (b)24/7
Explanation: AB=24cm and BC = 7cm
Tan C = Opposite side/Adjacent side
Tan C=24/7
4. Answer: (b) 1
Explanation: tan 60° = √3 and cot 30° = √3
Hence, tan 60°/cot 30° = √3/√3 = 1
5. Answer: (a)sin\(^2\)A
Explanation: We know, by trigonometry identities,
sin\(^2\)A+cos\(^2\)A = 1
1-cos\(^2\)A = sin\(^2\)A
6. Answer: (b) Same
Explanation: By trigonometry identities.
Sin (90°-A) = cos A [comes in the first quadrant of unit circle]
7. Answer: (d) sec x = cosec y
8. Answer: (b) 0
9. Answer: (b) 0
10. Answer: (c) trigonometric ratios of the angles
11. Answer: (b) cos 2β
Explanation: cos (α + β) = 0 = cos 90°
⇒ α + β = 90°
sin (α – β) = sin (α – β + β – β)
= sin (α + β – 2β)
= sin (90° – 2β) = cos 2β
12. Answer: (c) 1
Explanation: cos 9α = sin α
⇒ cos 9α = cos (90° – α)
⇒ 9α = 90° – α
⇒ 10α = 90°
⇒ α = 9°
∴ tan 5α = tan 5 × 9°
= tan 45° = 1
13. Answer: (b) 0
Explaination: sin (45° + θ) – cos (45° – θ)
= sin {90° – (45° – θ)} – cos (45° – θ)
= cos (45° – θ) – cos (45° – θ) = 0
14. Answer: (a) tan\(^2\)θ
Explanation: From trigonometric identity
1+ tan\(^2\)θ = sec\(^2\)θ
⇒sec\(^2\)θ – 1 = tan\(^2\)θ
15. Answer: (a) Are same
Explanation: Since from trigonometric identities,
cos(90° – θ) = sin θ
So, both represents the same value.
16. Answer: (c) 0
Explanation: Since
cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)
= cosec (75° + θ) – cosec [90° – (15° – θ)] – tan (55° + θ) + tan [90° – (35° – θ)]
= cosec (75° + θ) – cosec (75° + θ) – tan (55° + θ) + tan (55° + θ)
= 0
17. Answer: (c) 65° and 25°
18. Answer: (c) 45°
19. Answer: (c) 0°
Explanation: sin 2A = sin 0° = 0
2sin A = 2sin 0° = 0
20. Answer: (c) cot A
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