RD Sharma Solutions Class 11 Maths Chapter 10 - Sine And Cosine Formulae And Their Applications

Students preparing for their Class 11 Maths exams can access detailed RD Sharma Solutions for Chapter 10 – Sine and Cosine Formulae and Their Applications here. To excel in Mathematics and secure high marks, it is important for students to thoroughly practice all the exercises included in this chapter. This chapter focuses on important trigonometric identities and their connections with various elements of a triangle.

The solutions are thoughtfully prepared by subject matter experts to serve as a reliable learning resource for students. These step-by-step explanations help students clear their doubts quickly and confidently tackle examination questions.

RD Sharma Solutions Class 11 Maths Chapter 10 : Overview

Chapter 10 includes two major exercises that cover a range of important trigonometric concepts. The solutions have been created in alignment with the latest CBSE Class 11 syllabus, making it easier for students to follow standard methods and improve their problem-solving speed.

For convenient learning, students can download the RD Sharma Class 11 Maths Chapter 10 Solutions PDF through the direct links provided. Below is a quick summary of the key topics explained in this chapter:

• Law of Sines (Sine Rule)
• Law of Cosines
• Projection Formulae
• Napier’s Analogy (Law of Tangents)
• Area Calculation of a Triangle

These solutions are ideal for students who want clear, concise explanations and quick revision material. By regularly practicing these exercises, students can strengthen their understanding of trigonometry and perform well in their final exams.

RD Sharma Solutions Class 11 Maths - Sine And Cosine Formulae And Their Applications Question with Answers

State the Law of Sines:
Solution: a / sin A = b / sin B = c / sin C

State the Law of Cosines:
Solution: a² = b² + c² - 2bc cos A

In ΔABC, A = 30°, B = 60°, find C:
Solution: C = 180° - (30° + 60°) = 90°

In ΔABC, a = 7 cm, b = 9 cm, A = 45°, find B:
Solution: sin B = (9 × sin 45°) / 7 ≈ 0.91, B ≈ 65.5°

Find the area of ΔABC with a = 7 cm, b = 10 cm, C = 60°:
Solution: Area = 1/2 × 7 × 10 × sin 60° ≈ 30.31 cm²

In ΔABC, a = 8 cm, b = 6 cm, C = 60°, find c:
Solution: c = √(a² + b² - 2ab cos C) ≈ 7.21 cm

Prove the Projection Formula:
Solution: a = b cos C + c cos B

Find the area of an equilateral triangle with side 10 cm:
Solution: Area = (√3 / 4) × 10² ≈ 43.30 cm²

In a right triangle, a = 8 cm, b = 6 cm, A = 90°, find c:
Solution: c = √(64 + 36) = 10 cm

Prove cos A = (b² + c² - a²) / (2bc):
Solution: Directly follows from the Cosine Rule.

In ΔABC, A = 45°, B = 60°, a = 6 cm, find b:
Solution: b ≈ 7.35 cm using Sine Rule

Find the area of triangle with a = 5 cm, b = 6 cm, c = 7 cm:
Solution: Area ≈ 14.7 cm² using Heron’s Formula

Prove the Sine Rule:
Solution: a / sin A = b / sin B = c / sin C

In ΔABC, A = 60°, a = 12 cm, find R:
Solution: R = a / (2 sin A) = 4√3 cm

Find the area in ΔABC, A = 90°, a = 9 cm, b = 12 cm:
Solution: Area = 1/2 × 9 × 12 = 54 cm²

Prove cos² A + sin² A = 1:
Solution: This is the Pythagorean Identity.

In ΔABC, A = 30°, a = 10 cm, find R:
Solution: R = a / (2 sin A) = 10 cm

Verify the Sine Rule in an equilateral triangle:
Solution: True as all sides and angles are equal.

Show area = abc / (4R):
Solution: Follows from area formula combined with Sine Rule.

In ΔABC, a = 9 cm, b = 12 cm, A = 90°, find c:
Solution: c = √(a² + b²) = 15 cm

State and prove the Law of Sines.

Solution: The Law of Sines states:

a / sin A = b / sin B = c / sin C

Proof: In any triangle ABC, the area is given by:

Area = (1/2) bc sin A = (1/2) ac sin B = (1/2) ab sin C

By simplifying, we get a / sin A = b / sin B = c / sin C. Hence proved.

State and prove the Law of Cosines.

Solution: The Law of Cosines is:

a² = b² + c² - 2bc cos A

Proof: This formula is derived using the projection method and is a generalization of the Pythagoras Theorem.

In ΔABC, A = 60°, B = 45°, find angle C.

Solution:

C = 180° - (A + B) = 180° - (60° + 45°) = 75°

In ΔABC, a = 7 cm, b = 9 cm, A = 45°, find angle B.

Solution:

Using Sine Rule, sin B = (b × sin A) / a = (9 × sin 45°) / 7 ≈ 0.91

Therefore, B ≈ 65.5°

Find the area of triangle ABC if a = 6 cm, b = 8 cm, C = 60°.

Solution:

Area = (1/2) × a × b × sin C = (1/2) × 6 × 8 × sin 60°

Area ≈ 20.78 cm²

In ΔABC, a = 8 cm, b = 6 cm, C = 60°, find side c.

Solution:

By Cosine Rule, c² = a² + b² - 2ab cos C

c² = 64 + 36 - 48 = 52 → c = √52 ≈ 7.21 cm

Prove that in any triangle ABC, cos² A + sin² A = 1.

Solution: This is a standard trigonometric identity valid for all angles.

cos² A + sin² A = 1

Find the area of an equilateral triangle with side 10 cm.

Solution:

Area = (√3 / 4) × a² = (√3 / 4) × 100 = 25√3 ≈ 43.30 cm²

In ΔABC, a = 5 cm, b = 7 cm, C = 90°, find c.

Solution:

Since C = 90°, by Pythagoras theorem:

c² = a² + b² = 25 + 49 = 74 → c = √74 ≈ 8.60 cm

In ΔABC, A = 45°, a = 8 cm, find circumradius R.

Solution:

By Sine Rule, a / sin A = 2R → R = a / (2 × sin A)

R = 8 / (2 × 1/√2) = 4√2 ≈ 5.66 cm