Class 11: Maths Chapter 10 solutions. Complete Class 11 Maths Chapter 10 Notes.
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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
= 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 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
= (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 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.
9. For 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
Chapterwise RD Sharma Solutions for Class 11 Maths :
- Chapter 1–Sets
- Chapter 2–Relations
- Chapter 3–Functions
- Chapter 4–Measurement of Angles
- Chapter 5–Trigonometric Functions
- Chapter 6–Graphs of Trigonometric Functions
- Chapter 7–Values of Trigonometric Functions at Sum or Difference of Angles
- Chapter 8–Transformation Formulae
- Chapter 9–Values of Trigonometric Functions at Multiples and Submultiples of an Angle
- Chapter 10–Sine and Cosine Formulae and their Applications
- Chapter 11–Trigonometric Equations
- Chapter 12–Mathematical Induction
- Chapter 13–Complex Numbers
- Chapter 14–Quadratic Equations
- Chapter 15–Linear Inequations
- Chapter 16–Permutations
- Chapter 17–Combinations
- Chapter 18–Binomial Theorem
- Chapter 19–Arithmetic Progressions
- Chapter 20–Geometric Progressions
- Chapter 21–Some Special Series
- Chapter 22–Brief review of Cartesian System of Rectangular Coordinates
- Chapter 23–The Straight Lines
- Chapter 24–The Circle
- Chapter 25–Parabola
- Chapter 26–Ellipse
- Chapter 27–Hyperbola
- Chapter 28–Introduction to Three Dimensional Coordinate Geometry
- Chapter 29–Limits
- Chapter 30–Derivatives
- Chapter 31–Mathematical Reasoning
- Chapter 32–Statistics
- Chapter 33–Probability
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.