RD Sharma Solutions for Class 11 Maths Chapter 10 Sine and Cosine Formulae and their Applications

RD Sharma Solutions Class 11 Maths Chapter 10 – Free PDF Download

RD Sharma Solutions for Class 11 Maths Chapter 10 – Sine and Cosine Formulae and Their Applications are provided here for students to study and prepare for their final examination. To achieve a good score in Mathematics, students are required to practise all the questions of each and every exercise present in the chapter. In this chapter, we shall learn about some trigonometric relations with elements of a triangle. Experts at BYJU’S have formulated the RD Sharma Solutions with the intention of providing students with the best reference guide to erase their doubts immediately and score good marks in exams. 

RD Sharma Class 11 Solutions Chapter 10 – Sine and Cosine Formulae and Their Applications contains two exercises. The solutions curated by experts are in accordance with the CBSE syllabus for easy and quick calculations. Students can easily get the PDF of RD Sharma Solutions for Class 11 Maths Chapter 10 Sine and Cosine Formulae and Their Applications from the links given below. Now, let us have a look at the concepts discussed in this chapter.

• The law of sines or sine rule
• The law of cosines
• Projection formulae
• Napier’s analogy (law of tangents)
• Area of a triangle

RD Sharma Solutions for Class 11 Maths Chapter 10 – Sine and Cosine Formulae and Their Applications

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Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 10 – Sine and Cosine Formulae and Their Applications

Exercise 10.1 Solutions

Exercise 10.2 Solutions

Access answers to RD Sharma Solutions for Class 11 Maths Chapter 10 – Sine and Cosine Formulae and Their Applications

EXERCISE 10.1 PAGE NO: 10.12

1. If in a ∆ABC, ∠A = 45^o, ∠B = 60^o, and ∠C = 75^o; find the ratio of its sides.

Solution:

Given: In ∆ABC, ∠A = 45^o, ∠B = 60^o, and ∠C = 75^o

By using the sine rule, we get

a: b: c = 2: √6: (1+√3)

Hence, the ratio of the sides of the given triangle is a: b: c = 2: √6: (1+√3).

2. If in any ∆ABC, ∠C = 105^o, ∠B = 45^o, a = 2, then find b.

Solution:

Given: In ∆ABC, ∠C = 105^o, ∠B = 45^o, a = 2

We know, in a triangle,

∠A + ∠B + ∠C = 180°

∠A = 180° – ∠B – ∠C

Substituting the given values, we get

∠A = 180° – 45° – 105°

∠A = 30°

By using the sine rule, we get

3. In ∆ABC, if a = 18, b = 24 and c = 30 and ∠C = 90^o, find sin A, sin B and sin C.

Solution:

Given: In ∆ABC, a = 18, b = 24 and c = 30 and ∠C = 90^o

By using the sine rule, we get

In any triangle ABC, prove the following:

Solution:

By using the sine rule, we know,

= RHS

Hence, proved.

5. (a – b) cos C/2 = C sin (A – B)/2

Solution:

By using the sine rule we know,

= RHS

Hence, proved.

Solution:

By using the sine rule, we know,

Solution:

By using the sine rule, we know,

cos (A + B)/2 = cos (A/2 + B/2) = cos A/2 cos B/2 + sin A/2 sin B/2

cos (A – B)/2 = cos (A/2 – B/2) = cos A/2 cos B/2 – sin A/2 sin B/2

Substituting the above equations to equation (vi), we get,

Solution:

By using the sine rule, we know,

11. b sin B – c sin C = a sin (B – C)

Solution:

By using the sine rule, we know,

So, c = k sin C

Similarly, a = k sin A

And b = k sin B

We know,

Now, let us consider LHS.

b sin B – c sin C

Substituting the values of b and c to the above equation, we get

k sin B sin B – k sin C sin C = k (sin² B – sin² C) ……….(i)

We know,

Sin² B – sin² C = sin (B + C) sin (B – C),

Substituting the above values in equation (i), we get

k (sin² B – sin² C) = k (sin (B + C) sin (B – C)) [since, π = A + B + C ⇒ B + C = π –A]

The above equation becomes,

= k (sin (π –A) sin (B – C)) [since, sin (π – θ) = sin θ]

= k (sin (A) sin (B – C))

From the sine rule, a = k sin A, so the above equation becomes,

= a sin (B – C)

= RHS

Hence proved.

12. a² sin (B – C) = (b² – c²) sin A

Solution:

By using the sine rule, we know,

So, c = k sin C

Similarly, a = k sin A

And b = k sin B

We know,

Now, let us consider RHS.

(b² – c²) sin A …

Substituting the values of b and c in the above equation, we get

(b² – c²) sin A = [(k sin B)² – ( k sin C)²] sin A

= k² (sin² B – sin² C) sin A………. (i)

We know,

Sin² B – sin² C = sin (B + C) sin (B – C),

Substituting the above values in equation (i), we get

= k² (sin (B + C) sin (B – C)) sin A [since, π = A + B + C ⇒ B + C = π –A]

= k² (sin (π –A) sin (B – C)) sin A

= k² (sin (A) sin (B – C)) sin A [since, sin (π – θ) = sin θ]

Rearranging the above equation, we get

= (k sin (A))( sin (B – C)) (k sin A)

From the sine rule, a = k sin A, so the above equation becomes,

= a² sin (B – C)

= RHS

Hence, proved.

= RHS

Hence proved.

14. a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = 0

Solution:

By using the sine rule, we know,

a = k sin A, b = k sin B, c = k sin C

Let us consider LHS:

a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B)

Substituting the values of a, b, and c from the sine rule to the above equation, we get

a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = k sin A (sin B – sin C) + k sin B (sin C – sin A) + k sin C (sin A – sin B)

= k sin A sin B – k sin A sin C + k sin B sin C – k sin B sin A + k sin C sin A – k sin C sin B

Upon simplification, we get

= 0

= RHS

Hence, proved.

Upon simplification, we get,

= k² [sin A sin (B – C) + sin B sin (C – A) + sin C sin (A – B)]

We know sin (A – B) = sin A cos B – cos A sin B

Sin (B – C) = sin B cos C – cos B sin C

Sin (C – A) = sin C cos A – cos C sin A

So the above equation becomes,

= k² [sin A (sin B cos C – cos B sin C) + sin B (sin C cos A – cos C sin A) + sin C (sin A cos B – cos A sin B)]

= k² [sin A sin B cos C – sin A cos B sin C + sin B sin C cos A – sin B cos C sin A + sin C sin A cos B – sin C cos A sin B)]

Upon simplification, we get,

= 0

= RHS

Hence, proved.

EXERCISE 10.2 PAGE NO: 10.25

In any ∆ABC, prove the following:

1. In a ∆ABC, if a = 5, b = 6 and C = 60^o, show that its area is (15√3)/2 sq. units.

Solution:

Given:

In a ∆ABC, a = 5, b = 6 and C = 60^o

By using the formula,

Area of ∆ABC = 1/2 ab sin θ, where a and b are the lengths of the sides of a triangle and θ is the angle between sides.

So,

Area of ∆ABC = 1/2 ab sin θ

= 1/2 × 5 × 6 × sin 60^o

= 30/2 × √3/2

= (15√3)/2 sq. units

Hence proved.

2. In a ∆ABC, if a = √2, b = √3 and c = √5 show that its area is1/2 √6 sq. units.

Solution:

Given:

In a ∆ABC, a = √2, b = √3 and c = √5

By using the formulas,

We know, cos A = (b² + c² – a²)/2bc

By substituting the values, we get,

= [(√3)² + (√5)² – (√2)²] / [2 × √3 × √5]

= 3/√15

We know, Area of ∆ABC = 1/2 bc sin A

To find sin A,

Sin A = √(1 – cos² A) [by using trigonometric identity]

= √(1 – (3/√15)²)

= √(1- (9/15))

= √(6/15)

Now,

Area of ∆ABC = 1/2 bc sin A

= 1/2 × √3 × √5 × √(6/15)

= 1/2 √6 sq. units

Hence, proved.

3. The sides of a triangle are a = 4, b = 6 and c = 8, show that: 8 cos A + 16 cos B + 4 cos C = 17.

Solution:

Given:

Sides of a triangle are a = 4, b = 6 and c = 8

By using the formulas,

Cos A = (b² + c² – a²)/2bc

Cos B = (a² + c² – b²)/2ac

Cos C = (a² + b² – c²)/2ab

So now, let us substitute the values of a, b and c, and we get,

Cos A = (b² + c² – a²)/2bc

= (6² + 8² – 4²)/2×6×8

= (36 + 64 – 16)/96

= 84/96

= 7/8

Cos B = (a² + c² – b²)/2ac

= (4² + 8² – 6²)/2×4×8

= (16 + 64 – 36)/64

= 44/64

Cos C = (a² + b² – c²)/2ab

= (4² + 6² – 8²)/2×4×6

= (16 + 36 – 64)/48

= -12/48

= -1/4

Now, considering LHS,

8 cos A + 16 cos B + 4 cos C = 8 × 7/8 + 16 × 44/64 + 4 × (-1/4)

= 7 + 11 – 1

= 17

Hence, proved.

4. In a ∆ABC, if a = 18, b = 24, c = 30, find cos A, cos B and cos C

Solution:

Given:

Sides of a triangle are a = 18, b = 24 and c = 30

By using the formulas,

Cos A = (b² + c² – a²)/2bc

Cos B = (a² + c² – b²)/2ac

Cos C = (a² + b² – c²)/2ab

So now, let us substitute the values of a, b and c, and we get,

Cos A = (b² + c² – a²)/2bc

= (24² + 30² – 18²)/2×24×30

= 1152/1440

= 4/5

Cos B = (a² + c² – b²)/2ac

= (18² + 30² – 24²)/2×18×30

= 648/1080

= 3/5

Cos C = (a² + b² – c²)/2ab

= (18² + 24² – 30²)/2×18×24

= 0/864

= 0

∴ cos A = 4/5, cos B = 3/5, cos C = 0

5. For any ΔABC, show that b (c cos A – a cos C) = c² – a²

Solution:

Let us consider LHS:

b (c cos A – a cos C)

As LHS contain bc cos A and ab cos C, which can be obtained from cosine formulae.

From the cosine formula, we have

Cos A = (b² + c² – a²)/2bc

bc cos A = (b² + c² – a²)/2 … (i)

Cos C = (a² + b² – c²)/2ab

ab cos C = (a² + b² – c²)/2 … (ii)

Now, let us subtract equations (i) and (ii), and we get,

bc cos A – ab cos C = (b² + c² – a²)/2 – (a² + b² – c²)/2

= c² – a²

∴ b (c cos A – a cos C) = c² – a²

Hence, proved.

6. For any Δ ABC show that c (a cos B – b cos A) = a² – b²

Solution:

Let us consider LHS:

c (a cos B – b cos A)

As LHS contain ca cos B and cb cos A, which can be obtained from cosine formulae.

From the cosine formula, we have:

Cos A = (b² + c² – a²)/2bc

bc cos A = (b² + c² – a²)/2 … (i)

Cos B = (a² + c² – b²)/2ac

ac cos B = (a² + c² – b²)/2 … (ii)

Now, let us subtract equation (ii) from (i), and we get,

ac cos B – bc cos A = (a² + c² – b²)/2 – (b² + c² – a²)/2

= a² – b²

∴ c (a cos B – b cos A) = a² – b²

Hence, proved.

7. For any Δ ABC show that
2 (bc cos A + ca cos B + ab cos C) = a² + b² + c²

Solution:

Let us consider LHS/

2 (bc cos A + ca cos B + ab cos C)

As LHS contain 2ca cos B, 2ab cos C and 2cb cos A, which can be obtained from cosine formulae.

From the cosine formula. we have:

Cos A = (b² + c² – a²)/2bc

2bc cos A = (b² + c² – a²) … (i)

Cos B = (a² + c² – b²)/2ac

2ac cos B = (a² + c² – b²)… (ii)

Cos C = (a² + b² – c²)/2ab

2ab cos C = (a² + b² – c²) … (iii)

Now let us add equations (i), (ii) and (ii), and we get,

2bc cos A + 2ac cos B + 2ab cos C = (b² + c² – a²) + (a² + c² – b²) + (a² + b² – c²)

Upon simplification, we get,

= c² + b² + a²

2 (bc cos A + ac cos B + ab cos C) = a² + b² + c²

Hence, proved.

8. For any Δ ABC show that
(c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C

Solution:

Let us consider LHS:

(c² – a² + b²), (a² – b² + c²), (b² – c² + a²)

We know the sine rule in Δ ABC.

As LHS contain (c² – a² + b²), (a² – b² + c²) and (b² – c² + a²), which can be obtained from cosine formulae.

From the cosine formula, we have

Cos A = (b² + c² – a²)/2bc

2bc cos A = (b² + c² – a²)

Let us multiply both sides by tan A, and we get,

2bc cos A tan A = (b² + c² – a²) tan A

2bc cos A (sin A/cos A) = (b² + c² – a²) tan A

2bc sin A = (b² + c² – a²) tan A … (i)

Cos B = (a² + c² – b²)/2ac

2ac cos B = (a² + c² – b²)

Let us multiply both sides by tan B, and we get,

2ac cos B tan B = (a² + c² – b²) tan B

2ac cos B (sin B/cos B) = (a² + c² – b²) tan B

2ac sin B = (a² + c² – b²) tan B … (ii)

Cos C = (a² + b² – c²)/2ab

2ab cos C = (a² + b² – c²)

Let us multiply both sides by tan C, and we get,

2ab cos C tan C = (a² + b² – c²) tan C

2ab cos C (sin C/cos C) = (a² + b² – c²) tan C

2ab sin C = (a² + b² – c²) tan C … (iii)

As we are observing that sin terms are involved, so let’s use the sine formula.

From the sine formula we have,

Let us multiply abc to each of the expressions, and we get,

bc sin A = ac sin B = ab sin C

2bc sin A = 2ac sin B = 2ab sin C

∴ From equations (i), (ii) and (iii), we have,

(c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C

Hence proved.

9. For any Δ ABC, show that:

Solution:

Let us consider LHS:

We can observe that we can get the terms c – b cos A and b – c cos A from the projection formula.

From the projection formula, we get,

c = a cos B + b cos A

c – b cos A = a cos B …. (i)

And,

b = c cos A + a cos C

b – c cos A = a cos C …. (ii)

Dividing equation (i) by (ii), we get,

= RHS

Hence proved.