RD Sharma Class 10 Solutions Chapter 5 - Trigonometric Ratios

RD Sharma Class 10 Solutions Chapter 5 - Trigonometric Ratios: Trigonometry is an important part of Class 10 mathematics, and understanding the basics starts with learning about trigonometric ratios. The RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios provides a strong foundation for this topic. This chapter explains how to find the values of the trigonometric ratios such as sine, cosine, and tangent for different angles. With the help of RD Sharma Class 10 Solutions, students can easily solve all the problems and questions given in the textbook.

The RD Sharma Class 10 Chapter 5 focuses on the definition and calculation of trigonometric ratios for right-angled triangles. Important formulas and step-by-step examples are given in the RD Sharma Class 10 Trigonometric Ratios Solutions to make concepts clear. These solutions follow the latest CBSE Class 10 Maths syllabus and are also aligned with NCERT Solutions for Class 10 Maths for complete exam preparation.

Using the Class 10 RD Sharma Trigonometric Ratios chapter, students learn to apply trigonometric ratios in various mathematical problems and also prepare for higher-level maths. The RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios are presented in a simple language so every student can understand and practice well.

To access all the solved questions and formulas in a handy format, be sure to read the full article and download the Class 10 Chapter 5 PDF with complete solutions for better learning and quick revision.

RD Sharma Class 10 Chapter 5 Trigonometric Ratios PDF with Solutions

The RD Sharma Class 10 Solutions 2025-26 PDF for Chapter 5 is a very helpful resource for students learning Trigonometric Ratios. This chapter is important for building a strong base in trigonometry, which is useful for Class 10 board exams and higher studies too. The RD Sharma Class 10 Trigonometric Ratios PDF explains how to find sine, cosine, and tangent values in a easy step-by-step method.

Sometimes, students find trigonometry confusing, but the RD Sharma Class 10 Solutions 2025-26 PDF makes it clear with simple steps and examples. You can also get the RD Sharma Class 10 Trigonometric Ratios PDF download to study offline anytime you want.

This chapter follows the latest CBSE board guidelines and helps you solve all types of questions easily. Read the full article below to download the Class 10 Chapter 5 Trigonometric Ratios PDF with solutions for better practice.

Important Trigonometric Ratios Questions (with Solutions)

I. Basic Ratio Conversion

1. If sin A = 2/3, find all other trigonometric ratios of angle A.

Solution:
Let the triangle sides be: perpendicular (opposite) = 2, hypotenuse = 3.
Base = √(3² - 2²) = √5.

• sin A = 2/3
• cos A = √5 / 3
• tan A = 2 / √5
• cot A = √5 / 2
• sec A = 3 / √5
• cosec A = 3/2

2. If cos A = 4/5, find all other trigonometric ratios.

Solution:
Base = 4, Hypotenuse = 5 → Perpendicular = 3.

• sin A = 3/5
• tan A = 3/4
• cot A = 4/3
• sec A = 5/4
• cosec A = 5/3

3. If tan θ = 5/12, find sin θ, cos θ, sec θ, cosec θ, cot θ.

Solution:
Perpendicular = 5, Base = 12 → Hypotenuse = 13.

• sin θ = 5/13
• cos θ = 12/13
• sec θ = 13/12
• cosec θ = 13/5
• cot θ = 12/5

II. Solving Triangle Sides

4. In ΔABC, right-angled at B, AB = 24cm, BC = 7cm. Find sin A, cos A.

Solution:
AC = √(24² + 7²) = 25

• sin A = 7/25
• cos A = 24/25

5. In the same ΔABC, find sin C, cos C.

Solution:

• sin C = 24/25
• cos C = 7/25

III. Given One Ratio (Complex)

6. If cot A = 7/8, evaluate

1. ((1+sin A)(1−sin A)) / ((1+cos A)(1−cos A))
2. cot²A

Solutions:
(i) (1−sin² A)/(1−cos² A) = (cos² A)/(sin² A) = (cot A)² = (7/8)² = 49/64
(ii) cot²A = 49/64

7. If sin θ = 11/15, find all other trigonometric ratios.

Solution:
Opposite = 11, Hypotenuse = 15 → Base = √(225 − 121) = 2√26.

• cos θ = 2√26 / 15
• tan θ = 11 / 2√26
• cot θ = 2√26 / 11
• sec θ = 15 / 2√26
• cosec θ = 15/11

IV. Application/Values & Identities

8. Evaluate: sin 45°·sin 30° + cos 45°·cos 30°

Solution:
sin 45° = 1/√2, sin 30° = 1/2, cos 45° = 1/√2, cos 30° = √3/2
= (1/√2 × 1/2) + (1/√2 × √3/2) = (1/2√2) + (√3/2√2) = (1 + √3) / 2√2

9. Evaluate: sin 60°·cos 30° + cos 60°·sin 30°

Solution:
sin 60° = √3/2, cos 30° = √3/2, cos 60° = 1/2, sin 30° = 1/2
= (√3/2 × √3/2) + (1/2 × 1/2) = (3/4) + (1/4) = 1

10. Prove: sin θ·sin(90°–θ) – cos θ·cos(90°–θ) = 0

Solution:
sin(90°–θ) = cos θ, cos(90°–θ) = sin θ
LHS = sin θ·cos θ – cos θ·sin θ = 0, hence proved.

11. If tan θ = a/b, find (cos θ + sin θ) / (cos θ – sin θ)

Solution:
sin θ = a / √(a² + b²), cos θ = b / √(a² + b²)
Expression = (b + a) / (b - a)

V. Special Values & Short Calculations

12. Find value: sin²30° + sin²45° + sin²60° + sin²90°

Solution:
= (1/2)² + (1/√2)² + (√3/2)² + (1)² = 1/4 + 1/2 + 3/4 + 1 = 2.5

13. Evaluate: tan²30° + tan²45° + tan²60°

Solution:
tan 30° = 1/√3, tan 45° = 1, tan 60° = √3
Sum = (1/3) + 1 + 3 = 4.333…

14. If tan θ = 4/3, find (4 – tan θ) / (2 + tan θ)

Solution:
Substitute tan θ:
(4 – 4/3) / (2 + 4/3) = (12 – 4) / (6 + 4) = 8/10 = 4/5

15. If sec θ = 13/5, show that tan θ = 12/5

Solution:
Hypotenuse = 13, Base = 5 ⇒ Perpendicular = 12.
tan θ = 12/5

16. If cos θ = 12/13, prove sin θ(1 – tan θ) = 35/156

Solution:
sin θ = 5/13, tan θ = 5/12
(5/13)(1 – 5/12) = (5/13 × 7/12) = 35/156

17. If sin θ = √3/2, find cos θ, tan θ, sec θ

Solution:
sin θ = √3/2, so base = 1, hypotenuse = 2
cos θ = 1/2, tan θ = √3/1 = √3, sec θ = 2/1 = 2

18. In ΔPQR, right at Q, PQ = 4cm, RQ = 3cm. Find sin P, sin R, sec P, sec R.

Solution:
PR = 5
sin P = 3/5, sin R = 4/5, sec P = 5/4, sec R = 5/3

VI. Express in Terms of Ratios

19. Express cos 75° + cot 75° in terms of angles between 0° and 30°.

Solution:
cos 75° = sin 15°, cot 75° = tan 15°
Answer: sin 15° + tan 15°

20. If sin 3A = cos(A – 26°), where 3A acute, find value of A.

Solution:
sin 3A = sin(90° – (A – 26°)) ⇒ 3A = 90° – A + 26°
4A = 116° ⇒ A = 29°

RD Sharma Class 10 Solutions Chapter 5 - Trigonometric Ratios PDF Overview

The RD Sharma Class 10 Solutions Chapter 5 PDF is a great resource to help students learn and understand trigonometric ratios easily and quickly. This chapter is very important in the Class 10 maths syllabus and is a key topic in geometry and trigonometry. It teaches how to calculate the basic trigonometric ratios like sine, cosine, and tangent, which are used to solve problems involving right-angled triangles. Understanding these ratios is useful not only for exams but also for further studies in maths.

Main topics covered in Chapter 5:

• Introduction to trigonometric ratios (sin, cos, tan)
• Finding ratios for different angles
• Relationships between ratios
• Applying trigonometric ratios in problems
• Using trigonometry to solve real-life questions

The RD Sharma Class 10 Trigonometric Ratios Solutions provide step-by-step explanations, easy-to-understand methods, and useful tips for every question. Many students find trigonometry tricky, but this PDF makes it simple with clear solutions and examples.

Why choose RD Sharma Class 10 Solutions Chapter 5 PDF?

• Every question is solved clearly and simply
• Ideal for homework, practice, and exam preparation
• Important formulas and concepts are explained with examples
• You can download the PDF and study without internet
• Free access and helpful even if you make mistakes while solving

Advantages of practicing with RD Sharma Class 10 Chapter 5 - Trigonometric Ratios

• Builds strong foundation in trigonometry for higher studies
• Improves problem-solving skills with many practice questions
• Fully aligned with the latest CBSE Class 10 syllabus
• Helps improve speed and accuracy in exams
• Detailed stepwise solutions help avoid confusion
• Important for competitive exams like JEE and NEET preparation

Using the RD Sharma Class 10 Solutions Chapter 5 PDF regularly will help you master trigonometric ratios and boost your confidence in maths exams.