MCQ Questions | Class 10 Maths Chapter 8 | Introduction to Trigonometry with Answers

MCQ | Introduction to Trigonometry Class 10 | Sat C
Check the below NCERT MCQ Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Answers Pdf free download. MCQ Questions for Class 10 Maths with Answers were prepared based on the latest exam pattern. We have Provided Introduction to Trigonometry Class 10 Maths MCQs Questions with Answers to help students understand the concept very well.
Objective Question | NCERT Maths Class 10
Book: | National Council of Educational Research and Training (NCERT) |
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Board: | Central Board of Secondary Education (CBSE) |
Class: | 10th |
Subject: | Maths |
Chapter: | 8 |
Chapters Name: | Introduction to Trigonometry |
Medium: | English |
Ncert Solutions for Class 10 Maths | Objective Type Questions
You can refer to NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.
Introduction to Trigonometry | Class 10 Maths | NCERT Solutions
Q1. | In the given figure, if AB = 14 cm, then the value of tan B is: |
A. 1 – A, 2 – C, 3 – B |
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B. 1 – C, 2 – A, 3 – D |
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C. 1 – B, 2 – A, 3 – E |
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D. 1 – B, 2 – D, 3 – A |
Ans: 1 – B, 2 – D, 3 – A
Explaination: (4) definition of trigonometric ratios.
Explaination: (4) definition of trigonometric ratios.
Q2. | If 0° < θ < 90°, then sec 0 is |
A. >1 |
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B. < 1 |
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C. =1 |
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D. 0 |
Ans: >1
∵ sce θ > 1.
Explaination:
∵ sec θ = \(\frac{1}{\cos \theta}\)∵ sce θ > 1.
Q3. | If cos (α + β) = 0, then sin (α – β) can be reduced to |
A. cos β |
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B. cos 2β |
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C. sin α |
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D. sin 2α |
Ans: cos 2β
⇒ α + β = 90°
sin (α – β) = sin (α – β + β – β)
= sin (α + β – 2β)
= sin (90° – 2β) = cos 2β
Explaination:
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
Q4. | If cos 9α = sin a and 9α < 90°, then the value of tan 5α is |
A. \(\frac{1}{\sqrt{3}}\) |
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B. √3 |
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C. 1 |
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D. 0 |
Ans: 1
⇒ cos 9α = cos (90° – α)
⇒ 9α = 90° – α
⇒ 10α = 90°
⇒ α = 9°
∴ tan 5α = tan 5 × 9°
= tan 45° = 1
Explaination:
cos 9α = sin α⇒ cos 9α = cos (90° – α)
⇒ 9α = 90° – α
⇒ 10α = 90°
⇒ α = 9°
∴ tan 5α = tan 5 × 9°
= tan 45° = 1
Q5. | sin (45° + θ) – cos (45° – θ) is equal to |
A. 2 cos θ |
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B. 0 |
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C. 2 sin θ |
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D. 1 |
Ans:
= sin {90° – (45° – θ)} – cos (45° – θ)
= cos (45° – θ) – cos (45° – θ) = 0
Explaination:
sin (45° + θ) – cos (45° – θ)= sin {90° – (45° – θ)} – cos (45° – θ)
= cos (45° – θ) – cos (45° – θ) = 0
Q6. | The value of sin² 5° + sin² 10° + sin² 15° + … + sin² 90° is equal to |
A. 8 |
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B. 8.5 |
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C. 9 |
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D. 9.5 |
Ans: 9.5
= (sin² 5° + sin² 85°) + (sin² 10° + sin² 80°) + … + (sin² 40° + sin² 50°) + sin² 45° + sin² 90°
= (sin² 5° + cos² 5°) + (sin 10° + cos² 10°) + … + (sin² 40° + cos² 40°) + \(\left(\frac{1}{\sqrt{2}}\right)^{2}\) + 1
= 1 + 1 + 1 + … 8 times + \(\frac{1}{2}\) + 1
= 9\(\frac{1}{2}\) = 9.5
Explaination:
sin² 5° + sin² 10° + sin² 15° + ….. + sin² 90°= (sin² 5° + sin² 85°) + (sin² 10° + sin² 80°) + … + (sin² 40° + sin² 50°) + sin² 45° + sin² 90°
= (sin² 5° + cos² 5°) + (sin 10° + cos² 10°) + … + (sin² 40° + cos² 40°) + \(\left(\frac{1}{\sqrt{2}}\right)^{2}\) + 1
= 1 + 1 + 1 + … 8 times + \(\frac{1}{2}\) + 1
= 9\(\frac{1}{2}\) = 9.5
Q7. | The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is |
A. -1 |
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B. 0 |
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C. 1 |
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D. \(\frac{3}{2}\) |
Ans: 0
= cosec {90° – (15° – θ)} – sec (15° – θ) – tan {90° – (35° – θ)} + cot (35° – θ)
= sec (15° – θ) – sec (15° – θ) – cot (35° – θ) + cot (35° – θ) = 0
Explaination:
cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)= cosec {90° – (15° – θ)} – sec (15° – θ) – tan {90° – (35° – θ)} + cot (35° – θ)
= sec (15° – θ) – sec (15° – θ) – cot (35° – θ) + cot (35° – θ) = 0
Q8. | sin (90° – A) = |
A. sin A |
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B. tan A |
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C. cos A |
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D. cosec A |
Ans: cos A
Q9. | If sin x + cosec x = 2, then sin19x + cosec20x = |
A. 219 |
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B. 220 |
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C. 2 |
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D. 239 |
Ans: 2
⇒ sin x + \(\frac{1}{sin x}\) = 2
⇒ sin² x + 1 = 2 sin x
⇒ (sin x – 1)² = 0 => sin x = 1 => cosec x = 1
∴ sin19 x + cosec20 x = 1 + 1 = 2
Explaination:
sin x + cosec x = 2⇒ sin x + \(\frac{1}{sin x}\) = 2
⇒ sin² x + 1 = 2 sin x
⇒ (sin x – 1)² = 0 => sin x = 1 => cosec x = 1
∴ sin19 x + cosec20 x = 1 + 1 = 2
CBSE Solutions | Ncert Solutions Maths Class 10
NCERT Solutions | 10th Maths Guide
NCERT | Class 10 Maths Solution
Ncert Solutions for Class 10 Maths
Ncert Solutions for Class 10 Maths Chapter 1
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MCQ Questions for Class 10 Maths Chapter 1 Real Numbers Sat A
MCQ Questions for Class 10 Maths Chapter 1 Real Numbers Sat B
MCQ Questions for Class 10 Maths Chapter 1 Real Numbers Sat C
MCQ Questions for Class 10 Maths Chapter 1 Real Numbers Sat D
Ncert Solutions for Class 10 Maths Chapter 2
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MCQ Questions for Class 10 Maths Chapter 2 Polynomials Sat A
MCQ Questions for Class 10 Maths Chapter 2 Polynomials Sat B
MCQ Questions for Class 10 Maths Chapter 2 Polynomials Sat C
Ncert Solutions for Class 10 Maths Chapter 3
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MCQ Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Sat A
MCQ Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Sat B
MCQ Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Sat C
Ncert Solutions for Class 10 Maths Chapter 4
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MCQ Questions for Class 10 Maths Chapter 4 Quadratic Equations Sat A
MCQ Questions for Class 10 Maths Chapter 4 Quadratic Equations Sat B
MCQ Questions for Class 10 Maths Chapter 4 Quadratic Equations Sat C
Ncert Solutions for Class 10 Maths Chapter 5
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MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions Sat A
MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions Sat B
MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions Sat C
Ncert Solutions for Class 10 Maths Chapter 6
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MCQ Questions for Class 10 Maths Chapter 6 Triangles Sat A
MCQ Questions for Class 10 Maths Chapter 6 Triangles Sat B
Ncert Solutions for Class 10 Maths Chapter 7
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MCQ Questions for Class 10 Maths Chapter 7 Coordinate Geometry Sat A
MCQ Questions for Class 10 Maths Chapter 7 Coordinate Geometry Sat B
Ncert Solutions for Class 10 Maths Chapter 8
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MCQ Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry Sat A
MCQ Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry Sat B
MCQ Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry Sat C
Ncert Solutions for Class 10 Maths Chapter 9
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MCQ Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Sat A
MCQ Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Sat B
Ncert Solutions for Class 10 Maths Chapter 10
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MCQ Questions for Class 10 Maths Chapter 10 Circles Sat A
MCQ Questions for Class 10 Maths Chapter 10 Circles Sat B
Ncert Solutions for Class 10 Maths Chapter 11
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MCQ Questions for Class 10 Maths Chapter 11 Constructions Sat A
MCQ Questions for Class 10 Maths Chapter 11 Constructions Sat B
Ncert Solutions for Class 10 Maths Chapter 12
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MCQ Questions for Class 10 Maths Chapter 12 Areas Related to Circles Sat A
MCQ Questions for Class 10 Maths Chapter 12 Areas Related to Circles Sat B
Ncert Solutions for Class 10 Maths Chapter 13
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MCQ Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes Sat A
MCQ Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes Sat B
MCQ Questions for Class 10 Maths Chapter 13 Surface Areas and Volumes Sat C
Ncert Solutions for Class 10 Maths Chapter 14
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MCQ Questions for Class 10 Maths Chapter 14 Statistics Sat A
MCQ Questions for Class 10 Maths Chapter 14 Statistics Sat B
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