Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry
Important Questions for Class 10 Maths Chapter 8 Trigonometry - Explore detailed answers and step by step solutions which will help you ace your Class 10 Math examination.
Ananya Gupta18 Jan, 2024
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Important Questions for Class 10 Maths Chapter 8: Introduction to Trigonometry in Class 10 have been compiled to aid students aiming for high scores in their 2023-2024 board exams. These questions, available for download in PDF format on our website, adhere to the latest CBSE syllabus and align with the NCERT book.
Created after a detailed analysis of previous years' question papers, exam trends, and the latest sample papers, these questions are invaluable tools for exam preparation. By solving these questions, students can familiarize themselves with the diverse types of problems that may be presented in the final exam. Students can conveniently learn and solve these problems offline by downloading the Trigonometry Class 10 questions PDF. This resource is designed to provide comprehensive coverage of the trigonometry concept introduced in Chapter 8, elucidating the relationships between angles and sides of a triangle. The questions span various difficulty levels and are accompanied by complete explanations, ensuring a thorough understanding of trigonometric formulas and their application in solving numerical problems.CBSE Important Questions for Class 10 Maths
The six primary trigonometric ratios—sine, cosine, tangent, secant, cosecant, and cotangent—form the foundation of the trigonometry concept explored in this chapter. The questions not only test the application of these ratios but also help students grasp the underlying principles. Expertly solved by our teachers, these important questions serve as a valuable resource for swift comprehension. Students can access Class 10 Maths Chapter 8 Introduction to Trigonometry Multiple-Choice Questions (MCQs) to further enhance their understanding and practice. For additional practice, a set of supplementary questions on Chapter 8 is also provided, allowing students to reinforce their learning and hone their problem-solving skills. Overall, these important questions are crafted to empower students in their preparation for the final examination, offering a holistic understanding of trigonometry in Class 10 Mathematics.Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry Overview
In this article, we delve into a comprehensive discussion of various important questions from Chapter 8, "Introduction to Trigonometry," in the CBSE Class 10 syllabus for the academic year 2023-24. Students preparing for CBSE Class 10 Mathematics Standard and Mathematics Basic examination must prepare these questions very well in order to get great scores. Expertly solved by teachers, these important questions come with detailed explanations, facilitating a step-by-step understanding of the solutions. In addition to the core set of questions, supplementary problems are provided, allowing students to delve deeper into the chapter and practice further to strengthen their problem-solving skills. This collection of important questions aims not only to boost students' examination preparedness but also to foster a holistic understanding of trigonometry in Class 10 Mathematics. By engaging with these questions, students can gain confidence, proficiency, and a thorough grasp of the concepts essential for success in the upcoming board exams.Class 10 Maths Chapter 8 Important Questions and Answers
Q.1: Evaluate 2 tan 2 45° + cos 2 30° – sin 2 60°. Solution: Since we know, tan 45° = 1 cos 30° = √3/2 sin 60° = √3/2 Therefore, putting these values in the given equation: 2(1) 2 + (√3/2) 2 – (√3/2) 2 = 2 + 0 = 2 Q.2: If tan θ + cot θ = 5, find the value of tan2θ + cotθ. (2012)Solution:
tan θ + cot θ = 5 … [Given tan2θ + cot2θ + 2 tan θ cot θ = 25 … [Squaring both sides tan2θ + cot2θ + 2 = 25 ∴ tan2θ + cot2θ = 23 Q.3: If tan (A + B) =√3 and tan (A – B) =1/√3, 0° < A + B ≤ 90°; A > B, find A and B.Solution: Given,
tan (A + B) = √3 As we know, tan 60° = √3 Thus, we can write; ⇒ tan (A + B) = tan 60° ⇒(A + B) = 60° …… (i) Now again given; tan (A – B) = 1/√3 Since, tan 30° = 1/√3 Thus, we can write; ⇒ tan (A – B) = tan 30° ⇒(A – B) = 30° ….. (ii) Adding the equation (i) and (ii), we get; A + B + A – B = 60° + 30° 2A = 90° A= 45° Now, put the value of A in eq. (i) to find the value of B; 45° + B = 60° B = 60° – 45° B = 15° Therefore A = 45° and B = 15° Q.4: If tan α = 3 –√ and tan β = 1 3 √ ,0 < α, β < 90°, find the value of cot (α + β). (2012)Solution:
tan α = 3 –√ = tan 60° …(i) tan β = 1 3 √ = tan 30° …(ii) Solving (i) & (ii), α = 60° and β = 30° ∴ cot (α + β) = cot (60° + 30°) = cot 90° = 0 Q.5: If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.Solution: Given,
tan 2A = cot (A – 18°) As we know by trigonometric identities, tan 2A = cot (90° – 2A) Substituting the above equation in the given equation, we get; ⇒ cot (90° – 2A) = cot (A – 18°) Therefore, ⇒ 90° – 2A = A – 18° ⇒ 108° = 3A A = 108° / 3 Hence, the value of A = 36° Q.6: If sec θ + tan θ = 7, then evaluate sec θ – tan θ. (2017OD)Solution:
We know that, sec 2 θ – tan 2 θ = 1 (sec θ + tan θ) (sec θ – tan θ) = 1 (7) (sec θ – tan θ) = 1 …[sec θ + tan θ = 7; (Given) ∴ sec θ – tan θ = 1 7 Q.7: Show that : (i) tan 48° tan 23° tan 42° tan 67° = 1 (ii) cos 38° cos 52° – sin 38° sin 52° = 0Solution:
(i) tan 48° tan 23° tan 42° tan 67° We can also write the above given tan functions in terms of cot functions, such as; tan 48° = tan (90° – 42°) = cot 42° tan 23° = tan (90° – 67°) = cot 67° Hence, substituting these values, we get = cot 42° cot 67° tan 42° tan 67° = (cot 42° tan 42°) (cot 67° tan 67°) = 1 × 1 [since cot A.tan A = 1] = 1 (ii) cos 38° cos 52° – sin 38° sin 52° We can also write the given cos functions in terms of sin functions. cos 38° = cos (90° – 52°) = sin 52° cos 52°= cos (90° – 38°) = sin 38° Hence, putting these values in the given equation, we get; sin 52° sin 38° – sin 38° sin 52° = 0 Q.8: If cosec θ = 5 4 , find the value of cot θ. (2014)Solution:
We know that, cot 2 θ = cosec 2 θ – 1 = ( 5 4 ) 2 – 1 ⇒ 25 16 – 1 ⇒ 25 − 16 16 coť 2 θ = 9 16 i cot θ = 3 4 Q.9: If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.Solution:
As we know, for any given triangle, the sum of all its interior angles is equals to 180°. Thus, A + B + C = 180° ….(1) Now we can write the above equation as; ⇒ B + C = 180° – A Dividing by 2 on both the sides; ⇒ (B + C)/2 = (180° – A)/2 ⇒ (B + C)/2 = 90° – A/2 Now, put sin function on both sides. ⇒ sin (B + C)/2 = sin (90° – A/2) Since, sin (90° – A/2) = cos A/2 Therefore, sin (B + C)/2 = cos A/2 Q.10: What happens to value of cos when increases from 0° to 90°? (2015)Solution:
cos 0° = 1, cos 90° = 0 When θ increases from 0° to 90°, the value of cos θ decreases from 1 to 0. Q.11: If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1.Solution:
Given, sin θ + cos θ = √3 Squaring on both sides, (sin θ + cos θ) 2 = (√3) 2 sin 2 θ + cos 2 θ + 2 sin θ cos θ = 3 Using the identity sin 2 A + cos 2 A = 1, 1 + 2 sin θ cos θ = 3 2 sin θ cos θ = 3 – 1 2 sin θ cos θ = 2 sin θ cos θ = 1 sin θ cos θ = sin 2 θ + cos 2 θ ⇒ (sin 2 θ + cos 2 θ)/(sin θ cos θ) = 1 ⇒ [sin 2 θ/(sin θ cos θ)] + [cos 2 θ/(sin θ cos θ)] = 1 ⇒ (sin θ/cos θ) + (cos θ/sin θ) = 1 ⇒ tan θ + cot θ = 1 Hence proved. Q.12: Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.Solution:
cot 85° + cos 75° = cot (90° – 5°) + cos (90° – 15°) We know that cos(90° – A) = sin A and cot(90° – A) = tan A = tan 5° + sin 15° Q.13: What is the value of (cos 2 67° – sin 2 23°)?Solution:
(cos 2 67° – sin 2 23°) = cos 2 (90° – 23°) – sin 2 23° We know that cos(90° – A) = sin A = sin 2 23° – sin 2 23° = 0 Therefore, (cos 2 67° – sin 2 23°) = 1. Q.14: Express cot 75° + cosec 75° in terms of trigonometric ratios of angles between 0° and 30°. (2013)Solution:
cot 75° + cosec 75° = cot(90° – 15°) + cosec(90° – 15°) = tan 15° + sec 15° …[cot(90°-A) = tan A cosec(90° – A) = sec A Q.15: Prove that (sin A – 2 sin 3 A)/(2 cos 3 A – cos A) = tan A.Solution:
LHS = (sin A – 2 sin 3 A)/(2 cos 3 A – cos A) = [sin A(1 – 2 sin 2 A)]/ [cos A(2 cos 2 A – 1] Using the identity sin 2 θ + cos 2 θ = 1, = [sin A(sin 2 A + cos 2 A – 2 sin 2 A)]/ [cos A(2 cos 2 A – sin 2 A – cos 2 A] = [sin A(cos 2 A – sin 2 A)]/ [cos A(cos 2 A – sin 2 A)] = sin A/cos A = tan A = RHS Hence proved.Important Questions For Class 10 Maths Chapter 8 Benefits
Solving with important questions for Class 10 Maths Chapter 8 - Introduction to Trigonometry offers several benefits for students preparing for their board exams:Conceptual Reinforcement:
- The important questions cover key concepts of trigonometry, reinforcing understanding and application of the six primary trigonometric ratios.
Strategic Preparation:
- Created after analyzing previous year's question papers, exam trends, and sample papers, these questions offer a strategic approach to exam preparation.
Diverse Question Types:
- The set includes a variety of question types, allowing students to practice different problem-solving techniques and scenarios related to trigonometry.
Expert Solutions:
- Expertly solved questions come with detailed explanations, aiding students in understanding the step-by-step process of solving problems.
Comprehensive Exam Readiness:
- By solving these important questions, students can comprehensively prepare for the final exam, gaining confidence in their ability to tackle trigonometry-related problems.
| Important Questions for Class 10 Maths Chapter 8 |
| Chapter 1 Real Numbers |
| Chapter 2 Polynomials |
| Chapter 3 Linear Equations In Two Variables |
| Chapter 4 Quadratic Equations |
| Chapter 5 Arithmetic Progression |
| Chapter 6 Triangles |
| Chapter 7 Coordinate Geometry |
| Chapter 8 Introduction to Trigonometry |
| Chapter 9 Applications of Trigonometry |
| Chapter 10 Circles |
| Chapter 11 Constructions |
| Chapter 12 Areas Related to Circles |
| Chapter 13 Surface Areas and Volumes |
| Chapter 14 Statistics |
| Chapter 15 Probability |
| Download Hand Written Answer Sheets PDF’s | |
| Biology Hand Written Answer Sheets | Maths Hand Written Answer Sheets |
| Chemistry Hand Written Answer Sheets | Physics Hand Written Answer Sheets |
| English Hand Written Answer Sheets | |
Important Questions For Class 10 Maths Chapter 8 FAQs
What is Trigonometry?
Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles. It involves the study of trigonometric functions like sine, cosine, tangent, cosecant, secant, and cotangent.
Why is Trigonometry Important?
Trigonometry is essential in various fields, including physics, engineering, computer science, and architecture. It helps in measuring and analyzing angles and distances, making it a fundamental tool for solving real-world problems.
How are Trigonometric Functions Used in Real Life?
Trigonometric functions are used in real life for tasks like navigation, surveying, designing structures, analyzing mechanical motion, and understanding wave patterns. They provide a way to model and solve problems involving angles and distances.
What are Trigonometric Identities?
Trigonometric identities are equations that are true for all values of the variables where the functions are defined.
How Can I Solve Trigonometry Problems Involving Angles of Elevation and Depression?
To solve problems involving angles of elevation and depression, set up right-angled triangles and use trigonometric ratios. Pay attention to whether you are dealing with an angle of elevation (looking up) or an angle of depression (looking down).
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