RD Sharma Solutions for Class 10 Maths Chapter 5 Trigonometric Ratios - FREE PDF Download

RD Sharma Solutions for Class 10 Maths Chapter 5 – Trigonometric Ratios are available to help students thoroughly understand the concepts. Proper guidance is crucial for excelling in exams, and these solutions provide just that. Prepared by BYJU’S experts following the latest CBSE guidelines, the RD Sharma Solutions are designed in a simple, easy-to-understand language, ensuring students can easily grasp the material.

The PDF for RD Sharma Solutions Class 10 Chapter 5 Trigonometric Ratios is available here. The problems from the textbook have been solved by our Maths experts to help students tackle the questions without difficulty. By practicing these solutions, students can expect to achieve excellent results in their board exams.

Trigonometry, the science of measuring triangles, is a key part of this chapter. Students will learn about trigonometric ratios and their relationships, along with the trigonometric ratios of specific angles.

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RD Sharma Class 10 Chapter 5 PDF includes detailed solutions, examples, and extra questions to help you master real numbers and other topics. Click here to download the RD Sharma Class 10 Chapter 5 PDF.

Access Answers to RD Sharma Solutions for Class 10 Maths Chapter 5 Trigonometric Ratios

Question 1: If sin θ = 3/5, find the value of cos θ and tan θ.

Solution: Using the Pythagorean identity sin² θ + cos² θ = 1, 

(3/5)² + cos² θ = 1 ⟹ 9/25 + cos² θ = 1 ⟹ cos² θ = 16/25 ⟹ cos θ = 4/5.

To find tan θ, 

use tan θ = sin θ / cos θ 

= (3/5) ÷ (4/5) 

= 3/4.

Question 2: If tan θ = 2, find the value of sin θ and cos θ.

Solution:tan θ = 2 ⟹ sin θ = 2 cos θ.

Using the identity sin² θ + cos² θ = 1, 

(2 cos θ)² + cos² θ = 1 

= 4 cos² θ + cos² θ = 1 

= 5 cos² θ = 1 ⟹ cos θ = 1/√5. 

Therefore, sin θ = 2 × 1/√5 = 2/√5.

Question 3: Find the value of cot 45°.

Solution: cot 45° = 1 / tan 45° 

= 1 / 1 

= 1.

Question 4: If cos θ = 7/25, find the value of sec θ.

Solution: Using the identity sin² θ + cos² θ = 1, 

sin² θ + (7/25)² = 1 

= sin² θ + 49/625 = 1 

= sin² θ = 576/625 

= sin θ = 24/25.

Now, sec θ = 1 / cos θ 

= 1 / (7/25) 

= 25/7.

Question 5: If tan θ = 5/12, find sin θ and cos θ.

Solution: From tan θ = 5/12, we have sin θ = 5/13 and cos θ = 12/13 by using the Pythagorean identity.

Question 6: If cos θ = 7/25, find the value of sin θ.

Solution: Using the identity sin² θ + cos² θ = 1, 

sin² θ + (7/25)² = 1 

= sin² θ + 49/625 = 1 

= sin² θ = 576/625 

= sin θ = 24/25.

Question 7: Find the value of sin 30°.

Solution: sin 30° = 1/2.

Question 8: Find the value of cos 60°.

Solution: cos 60° = 1/2.

Question 9: If sin θ = 5/13, find tan θ.

Solution: Using the identity sin² θ + cos² θ = 1, 

(5/13)² + cos² θ = 1 

= 25/169 + cos² θ = 1 

= cos² θ = 144/169 

= cos θ = 12/13.

Now, tan θ = sin θ / cos θ 

= (5/13) ÷ (12/13) 

= 5/12.

Question 10: Find the value of sec 90°.

Solution: sec 90° = 1 / cos 90° = 1 / 0 (undefined).