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

Class 11: Maths Chapter 10 solutions. Complete Class 11 Maths Chapter 10 Notes.

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

RD Sharma 11th Maths Chapter 10, Class 11 Maths Chapter 10 solutions

EXERCISE 10.1 PAGE NO: 10.12

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

Solution:

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

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 = 105o, ∠B = 45o, a = 2, then find b.

Solution:

Given: In ∆ABC, ∠C = 105o, ∠B = 45o, 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 = 90o, find sin A, sin B and sin C.

Solution:

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

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 in 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 in 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 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

= k(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

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

= k(sin (π –A) sin (B – C)) sin A

= k(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 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, c from sine rule in 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 = 60o, show that its area is (15√3)/2 sq. units.

Solution:

Given:

In a ∆ABC, a = 5, b = 6 and C = 60o

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 60o

= 30/2 × 3/2

= (153)/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 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 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 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 equation (i) and (ii) 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 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) 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 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 equation (i), (ii) and (ii) 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 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 cosine formula we have:

Cos A = (b2 + c2 – a2)/2bc

2bc cos A = (b2 + c2 – a2)

Let us multiply both the sides by tan A 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 the sides by tan B 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 the sides by tan C 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 being involved so let’s use sine formula.

From sine formula we have,

Let us multiply abc to each of the expression we get,

bc sin A = ac sin B = ab sin C

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

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

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

Hence proved.

9For any Δ ABC show that:

Solution:

Let us consider LHS:

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

From 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.

RD Sharma Solutions for Class 11 Maths Chapter 10: Download PDF

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

Download PDF: RD Sharma Solutions for Class 11 Maths Chapter 10–Sine and Cosine Formulae and their Applications PDF

Chapterwise RD Sharma Solutions for Class 11 Maths :

About RD Sharma

RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.