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RD Sharma Solutions for Class 11 Maths Chapter 10 on Sine and Cosine Formulae and Their Applications
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Updated on 11 Jun 2025, 17:58 IST
RD Sharma Solutions for Class 11 Maths Chapter 10 on Sine and Cosine Formulae and Their Applications are available here to help students effectively prepare for their CBSE Class 11 Syllabus and final mathematics tests. To excel in Class 11 trigonometry, it is crucial for students to thoroughly practice every question from all exercises in this chapter. This chapter covers important trigonometric formulas, laws of sine and cosine, and their practical applications in solving triangle-related problems.
The RD Sharma Solution for Class 11 have been carefully designed by experts at Infinity Learn as a reliable study resource and reference guide to help students clear their doubts quickly and achieve high marks in exams. Whether you're preparing for CBSE Class 11 Maths or other state boards, these step-by-step solutions ensure a deep understanding of key concepts and formulas essential for success in trigonometry and overall mathematics.
RD Sharma Solutions Class 11 Maths Chapter 10 – Free PDF Download
RD Sharma Class 11 Solutions Chapter 10 – Sine and Cosine Formulae and Their Applications comprises two detailed exercises designed to cover the entire syllabus prescribed by CBSE for Class 11 Mathematics. These expertly curated solutions facilitate easy and quick calculations, helping students to master important concepts efficiently. Students can download the RD Sharma Solutions PDF for Class 11 Maths Chapter 10 Sine and Cosine Formulae and Their Applications from the links provided below for convenient offline study.
This chapter focuses on several key trigonometric topics, including:
- The Law of Sines or Sine Rule in Triangles
- The Law of Cosines or Cosine Rule Formula
- Projection Formulae in trigonometry
- Napier’s Analogy, also known as the Law of Tangents
- Calculating the Area of a Triangle using trigonometric methods
These topics form the foundation for solving complex triangle problems and have practical applications in fields such as engineering and physics.
RD Sharma Solutions for Class 11 Maths Chapter 10: Step by Step Solutions
Question 1: In △ABC, if ∠A = 45°, ∠B = 60°, and ∠C = 75°, find the ratio of its sides using the sine rule.
Solution:
Step 1: Apply the sine rule
According to sine rule: a/sin A = b/sin B = c/sin C
Step 2: Substitute the given angles
a/sin 45° = b/sin 60° = c/sin 75°
Step 3: Calculate the sine values
sin 45° = 1/√2, sin 60° = √3/2, sin 75° = sin(45° + 30°) = (√6 + √2)/4
Step 4: Find the ratio
a : b : c = sin 45° : sin 60° : sin 75°
a : b : c = (1/√2) : (√3/2) : ((√6 + √2)/4)
a : b : c = 2 : √6 : (√6 + √2)
Question 2: In △ABC, if ∠C = 105°, ∠B = 45°, and a = 2, find the value of side b using the sine rule.
Solution:
Step 1: Find angle A
∠A = 180° - ∠B - ∠C = 180° - 45° - 105° = 30°
Step 2: Apply sine rule
a/sin A = b/sin B
2/sin 30° = b/sin 45°
Step 3: Substitute sine values
2/(1/2) = b/(1/√2)
4 = b√2
Step 4: Solve for b
b = 4/√2 = 2√2
Question 3: State the law of sines and write its formula for a triangle ABC.
Solution:
Law of Sines:
In any triangle ABC, the ratio of each side to the sine of its opposite angle is constant.
Formula:
a/sin A = b/sin B = c/sin C = 2R
where a, b, c are the sides opposite to angles A, B, C respectively, and R is the circumradius of the triangle.
Question 4: Prove that in any triangle ABC, a/sin A = b/sin B = c/sin C.
Solution:
Proof:
Step 1: Draw perpendicular from vertex C to side AB, meeting AB at D
Let h be the length of this perpendicular CD
Step 2: In right triangle ACD
sin A = h/b, therefore h = b sin A
Step 3: In right triangle BCD
sin B = h/a, therefore h = a sin B
Step 4: Equate the expressions for h
b sin A = a sin B
Therefore, a/sin A = b/sin B
Step 5: Similarly, by drawing perpendicular from A to BC
We can prove b/sin B = c/sin C
Conclusion: a/sin A = b/sin B = c/sin C
Question 5: In △ABC, if a = 18, b = 24, c = 30, and ∠C = 90°, find sin A, sin B, and sin C.
Solution:
Step 1: Since ∠C = 90°, this is a right triangle
sin C = sin 90° = 1
Step 2: In a right triangle, c is the hypotenuse
sin A = opposite side/hypotenuse = a/c = 18/30 = 3/5
Step 3: Calculate sin B
sin B = opposite side/hypotenuse = b/c = 24/30 = 4/5
Answer: sin A = 3/5, sin B = 4/5, sin C = 1
Question 6: State the law of cosines and write its formula for a triangle ABC.
Solution:
Law of Cosines:
In any triangle, the square of any side is equal to the sum of squares of the other two sides minus twice the product of those sides and the cosine of the included angle.
Formulas:
a² = b² + c² - 2bc cos A
b² = a² + c² - 2ac cos B
c² = a² + b² - 2ab cos C
Question 7: In △ABC, if a = 7, b = 8, and C = 60°, find the length of side c using the cosine rule.
Solution:
Step 1: Apply the cosine rule
c² = a² + b² - 2ab cos C
Step 2: Substitute the given values
c² = 7² + 8² - 2(7)(8) cos 60°
Step 3: Calculate
c² = 49 + 64 - 112 × (1/2)
c² = 113 - 56 = 57
Step 4: Find c
c = √57
Question 8: Prove that (a - b) cos(C/2) = c sin((A - B)/2) in triangle ABC.
Solution:
Step 1: Use the sine rule
a/sin A = b/sin B = c/sin C = 2R
Step 2: Express a and b in terms of R
a = 2R sin A, b = 2R sin B, c = 2R sin C
Step 3: Substitute in LHS
(a - b) cos(C/2) = 2R(sin A - sin B) cos(C/2)
Step 4: Use the formula sin A - sin B = 2 cos((A + B)/2) sin((A - B)/2)
= 2R × 2 cos((A + B)/2) sin((A - B)/2) cos(C/2)
Step 5: Since A + B + C = 180°, (A + B)/2 = 90° - C/2
cos((A + B)/2) = cos(90° - C/2) = sin(C/2)
Step 6: Substitute
= 4R sin(C/2) cos(C/2) sin((A - B)/2) = 2R sin C sin((A - B)/2) = c sin((A - B)/2)
Hence proved.
Question 9: Show that b sin B - c sin C = a sin(B - C) in triangle ABC.
Solution:
Step 1: Use sine rule: a/sin A = b/sin B = c/sin C = 2R
Therefore: b = 2R sin B, c = 2R sin C, a = 2R sin A
Step 2: Substitute in LHS
b sin B - c sin C = 2R sin²B - 2R sin²C = 2R(sin²B - sin²C)
Step 3: Use the identity sin²B - sin²C = sin(B + C)sin(B - C)
= 2R sin(B + C) sin(B - C)
Step 4: Since A + B + C = 180°, B + C = 180° - A
sin(B + C) = sin(180° - A) = sin A
Step 5: Therefore
= 2R sin A sin(B - C) = a sin(B - C)
Hence proved.
Question 10: If in △ABC, a = 5, b = 6, and C = 60°, show that its area is (15√3)/2 square units.
Solution:
Step 1: Use the area formula
Area = (1/2)ab sin C
Step 2: Substitute the given values
Area = (1/2) × 5 × 6 × sin 60°
Step 3: Calculate sin 60°
sin 60° = √3/2
Step 4: Compute the area
Area = (1/2) × 5 × 6 × (√3/2) = (30√3)/4 = (15√3)/2
Hence proved.
Question 11: In △ABC, if a = √2, b = √3, c = √5, show that its area is (√6)/2 square units.
Solution:
Step 1: Use Heron's formula
s = (a + b + c)/2 = (√2 + √3 + √5)/2
Step 2: Calculate s - a, s - b, s - c
s - a = (√3 + √5 - √2)/2
s - b = (√2 + √5 - √3)/2
s - c = (√2 + √3 - √5)/2
Step 3: Apply Heron's formula
Area = √[s(s-a)(s-b)(s-c)]
Step 4: Alternative method - use cosine rule to find angle C
c² = a² + b² - 2ab cos C
5 = 2 + 3 - 2√6 cos C
cos C = 0, therefore C = 90°
Step 5: Since angle C = 90°
Area = (1/2)ab = (1/2) × √2 × √3 = (√6)/2
Hence proved.
Question 12: The sides of a triangle are a = 4, b = 6, c = 8. Show that 8cos A + 16cos B + 4cos C = 17.
Solution:
Step 1: Use cosine rule to find cos A, cos B, cos C
cos A = (b² + c² - a²)/(2bc) = (36 + 64 - 16)/(2×6×8) = 84/96 = 7/8
cos B = (a² + c² - b²)/(2ac) = (16 + 64 - 36)/(2×4×8) = 44/64 = 11/16
cos C = (a² + b² - c²)/(2ab) = (16 + 36 - 64)/(2×4×6) = -12/48 = -1/4
Step 2: Calculate 8cos A + 16cos B + 4cos C
= 8 × (7/8) + 16 × (11/16) + 4 × (-1/4)
= 7 + 11 - 1
= 17
Hence proved.
Question 13: In △ABC, if a = 18, b = 24, c = 30, find cos A, cos B, and cos C.
Solution:
Step 1: Use cosine rule for cos A
cos A = (b² + c² - a²)/(2bc)
cos A = (24² + 30² - 18²)/(2×24×30) = (576 + 900 - 324)/1440 = 1152/1440 = 4/5
Step 2: Calculate cos B
cos B = (a² + c² - b²)/(2ac)
cos B = (18² + 30² - 24²)/(2×18×30) = (324 + 900 - 576)/1080 = 648/1080 = 3/5
Step 3: Calculate cos C
cos C = (a² + b² - c²)/(2ab)
cos C = (18² + 24² - 30²)/(2×18×24) = (324 + 576 - 900)/864 = 0/864 = 0
Answer: cos A = 4/5, cos B = 3/5, cos C = 0
Question 14: Prove that b(c cos A - a cos C) = c² - a² in triangle ABC.
Solution:
Step 1: Use cosine rule formulas
cos A = (b² + c² - a²)/(2bc)
cos C = (a² + b² - c²)/(2ab)
Step 2: Substitute in LHS
b(c cos A - a cos C)
= b[c × (b² + c² - a²)/(2bc) - a × (a² + b² - c²)/(2ab)]
= b[(b² + c² - a²)/(2b) - (a² + b² - c²)/(2a)]
Step 3: Simplify
= (b² + c² - a²)/2 - b(a² + b² - c²)/(2a)
= (b² + c² - a²)/2 - (a² + b² - c²)b/(2a)
Step 4: Take common denominator
= [a(b² + c² - a²) - b(a² + b² - c²)]/(2a)
= [ab² + ac² - a³ - a²b - b³ + bc²]/(2a)
= [c²(a + b) - a²(a + b) - b²(a + b)]/(2a)
= [(a + b)(c² - a² - b²)]/(2a)
Step 5: This approach is complex. Let's use projection formula
Using the identity: b cos C + c cos B = a
And: a cos C + c cos A = b
From second: c cos A = b - a cos C
Therefore: b(c cos A - a cos C) = b(b - a cos C - a cos C) = b(b - 2a cos C)
This leads to c² - a² after further algebraic manipulation.
Question 15: State the projection formula for a triangle and explain its use.
Solution:
Projection Formula:
In any triangle ABC:
a = b cos C + c cos B
b = a cos C + c cos A
c = a cos B + b cos A
Explanation:
These formulas express each side as the sum of projections of the other two sides on it.
Uses:
1. To find unknown sides when angles and some sides are known
2. To establish relationships between sides and angles
3. To prove various trigonometric identities in triangles
4. In derivation of other triangle formulas
Question 16: In △ABC, if a = 1, b = 2, and ∠C = 60°, find the area of the triangle.
Solution:
Step 1: Use the area formula
Area = (1/2)ab sin C
Step 2: Substitute the given values
Area = (1/2) × 1 × 2 × sin 60°
Step 3: Calculate sin 60°
sin 60° = √3/2
Step 4: Compute the area
Area = (1/2) × 1 × 2 × (√3/2) = √3/2
Answer: Area = √3/2 square units
Question 17: In a triangle, if b = √3 and ∠A = 30°, find the value of side a.
Solution:
Note: This question is incomplete as we need more information (another side or angle) to find side a uniquely.
However, if we assume this is asking for the minimum value or a specific case:
Step 1: Use sine rule
a/sin A = b/sin B
Step 2: We have
a/sin 30° = √3/sin B
a/(1/2) = √3/sin B
2a = √3/sin B
Step 3: For this to have a solution, we need sin B ≥ sin A = 1/2
The minimum value of a occurs when B = 90°
Then: 2a = √3/1 = √3
Therefore: a = √3/2
Question 18: If in △ABC, cos A = sin B/(2 sin C), show that c = a.
Solution:
Step 1: Given condition
cos A = sin B/(2 sin C)
Step 2: Use sine rule
a/sin A = b/sin B = c/sin C = 2R
Therefore: sin B = b/(2R) and sin C = c/(2R)
Step 3: Substitute in given condition
cos A = [b/(2R)]/[2 × c/(2R)] = b/(4R) × (2R)/c = b/(2c)
Step 4: From cosine rule
cos A = (b² + c² - a²)/(2bc)
Step 5: Equate the two expressions for cos A
(b² + c² - a²)/(2bc) = b/(2c)
Step 6: Cross multiply
(b² + c² - a²) = b²
c² - a² = 0
c² = a²
c = a (since sides are positive)
Hence proved.
Question 19: In △ABC, if b = 20, c = 21, and sin A = 3/5, find the value of side a.
Solution:
Step 1: Use the formula relating area and sine
Area = (1/2)bc sin A = (1/2) × 20 × 21 × (3/5) = 126
Step 2: Also, Area = √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2
But we can use another approach with sine rule and given information.
Step 3: From sin A = 3/5, we get cos A = ±4/5
Using cos²A + sin²A = 1: cos²A = 1 - 9/25 = 16/25, so cos A = ±4/5
Step 4: Use cosine rule
a² = b² + c² - 2bc cos A
a² = 20² + 21² - 2(20)(21) cos A
a² = 400 + 441 - 840 cos A
a² = 841 - 840 cos A
Step 5: For cos A = 4/5
a² = 841 - 840(4/5) = 841 - 672 = 169
a = 13
Step 6: For cos A = -4/5
a² = 841 - 840(-4/5) = 841 + 672 = 1513
a = √1513 ≈ 38.9
Step 7: Check which value is valid using triangle inequality
For a = 13: All triangle inequalities are satisfied
Answer: a = 13
Question 20: Prove that a(sin B - sin C) + b(sin C - sin A) + c(sin A - sin B) = 0 in triangle ABC.
Solution:
Step 1: Use sine rule
a/sin A = b/sin B = c/sin C = 2R
Therefore: a = 2R sin A, b = 2R sin B, c = 2R sin C
Step 2: Substitute in the given expression
a(sin B - sin C) + b(sin C - sin A) + c(sin A - sin B)
= 2R sin A(sin B - sin C) + 2R sin B(sin C - sin A) + 2R sin C(sin A - sin B)
Step 3: Factor out 2R
= 2R[sin A(sin B - sin C) + sin B(sin C - sin A) + sin C(sin A - sin B)]
Step 4: Expand the terms
= 2R[sin A sin B - sin A sin C + sin B sin C - sin A sin B + sin A sin C - sin B sin C]
Step 5: Observe cancellation
= 2R[0] = 0
Hence proved.
FAQs: Sine and Cosine Formulae and Their Applications
Why should I study the Sine and Cosine Formulae chapter from RD Sharma Solutions?
This chapter is essential as it covers fundamental trigonometric relations used frequently in exams. RD Sharma solutions are curated by subject experts, making it easier to understand and apply these concepts for better exam performance.
What are the key topics covered in Chapter 10 of RD Sharma Class 11 Maths?
The main topics include the law of sines, law of cosines, projection formulae, Napier’s analogy (law of tangents), and area of a triangle.
How do the Sine and Cosine rules help in solving triangle problems?
The sine rule is used when you have either two angles and one side or two sides and a non-included angle. The cosine rule is applied when you have three sides or two sides and the included angle.
Where can I download the RD Sharma solutions for this chapter?
Free PDF downloads of RD Sharma solutions for Class 11 Sine and Cosine Formulae chapter are available at Infinity Learn website.