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Class 11 RD Sharma Solutions – Chapter 10 Sine and Cosine Formulae and Their Applications – Exercise 10.1 | Set 1

  • Last Updated : 16 May, 2021

Question 1: If in △ABC, ∠A=45°, ∠B=60°, and ∠C=75°, find the ratio of its sides.

Solution:

According to the sine rule

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\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda



Hence, we get

\frac{a}{sin \hspace{0.1cm}45\degree} = \frac{b}{sin \hspace{0.1cm}60\degree} = \frac{c}{sin \hspace{0.1cm}75\degree} = \lambda

Using the formula,

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

sin (45°+30°) = sin(45°) cos(30°) + cos(45°) sin(30°)

sin 75° =(\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}})(\frac{1}{2})

sin 75° =\frac{1}{2\sqrt{2}}(\sqrt{3}+1)

\frac{a}{\frac{1}{\sqrt{2}}} = \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{\frac{1}{2\sqrt{2}}(1+\sqrt{3})} = \lambda



Multiplying the denominator by 2√2, we get

\frac{a}{\frac{2\sqrt{2}}{\sqrt{2}}} = \frac{b}{\frac{2\sqrt{2}\sqrt{3}}{2}} = \frac{c}{\frac{2\sqrt{2}}{2\sqrt{2}}(1+\sqrt{3})}\\ \frac{a}{2} = \frac{b}{\sqrt{6}} = \frac{c}{(1+\sqrt{3})}

Hence, we get

a : b : c = 2 : √6 : (√3+1)

Question 2: If in △ABC, ∠C=105°, ∠B=45°, and a=2, then find b.

Solution:

In the given condition, ∠C=105° and ∠B=45°

And as we know that,

A+B+C = π (Sum of all angles in triangle is supplementary)

A = π-(B+C)

A = 180°-(45°+105°)



A = 30°

Now, according the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B}\\ \frac{2}{sin \hspace{0.1cm}30\degree} = \frac{b}{sin \hspace{0.1cm}45\degree}\\ \frac{2}{\frac{1}{2}} = \frac{b}{\frac{1}{\sqrt{2}}}\\ b = 4 \times \frac{1}{\sqrt{2}}

After rationalizing the denominator, we get

b = 4 \times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\\ b = 4\times \frac{\sqrt{2}}{2}\\ b = 2\sqrt{2}

Question 3: In △ABC, if a=18, b=24, and c=30 and ∠C=90°, find sin A, sin B, and sin C.

Solution:

In the given condition, a=18, b=24 and c=30 and ∠C=90°

Now, according the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{c}{sin \hspace{0.1cm}C}\\ sin \hspace{0.1cm}A=\frac{a(sin \hspace{0.1cm}C)}{c}\\ sin \hspace{0.1cm}A=\frac{18(sin \hspace{0.1cm}90\degree)}{30}\\ sin \hspace{0.1cm}A=\frac{18}{30}\\ sin \hspace{0.1cm}A=\frac{3}{5}

Also,



\frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C}\\ sin \hspace{0.1cm}B=\frac{b(sin \hspace{0.1cm}C)}{c}\\ sin \hspace{0.1cm}B=\frac{24(sin \hspace{0.1cm}90\degree)}{30}\\ sin \hspace{0.1cm}B=\frac{24}{30}\\ sin \hspace{0.1cm}B=\frac{4}{5}\\ sin \hspace{0.1cm}C=90\degree=1

Question 4: In △ABC, prove the following:

\frac{a-b}{a+b} = \frac{tan(\frac{A-B}{2})}{tan(\frac{A+B}{2})}

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering the LHS of the equation, we have

\frac{a-b}{a+b} = \frac{\lambda(sinA-sinB)}{\lambda(sinA+sinB)}

By using trigonometric formula,

sin A – sin B = 2 sin(\frac{A-B}{2}) cos(\frac{A+B}{2})

sin A + sin B = 2 sin(\frac{A+B}{2}) cos(\frac{A-B}{2})

LHS = \frac{2\hspace{0.1cm}sin(\frac{A-B}{2})cos(\frac{A+B}{2})}{2\hspace{0.1cm}sin(\frac{A+B}{2})cos(\frac{A-B}{2})}\\ = \frac{tan(\frac{A-B}{2})}{tan(\frac{A+B}{2})}\\ = RHS



As, LHS = RHS

Hence, proved !!

Question 5: In △ABC, prove the following:

(a-b)cos(\frac{C}{2}) = csin(\frac{A-B}{2})

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS, we have

(a-b) cos(\frac{C}{2}) = λ(sin A-sin B) cos(\frac{C}{2})

By using trigonometric formula,

sin A – sin B = 2 sin(\frac{A-B}{2}) cos(\frac{A+B}{2})

= λ(2 sin (\frac{A-B}{2}) cos (\frac{A+B}{2})) cos(\frac{C}{2})



A+B+C = π (Sum of all angles in triangle is supplementary)

= 2λ sin(\frac{A-B}{2}) cos(\frac{A+B}{2}) cos(\frac{\pi-(A+B)}{2})

= 2λ sin(\frac{A-B}{2}) cos(\frac{A+B}{2}) cos(\frac{\pi}{2}-(\frac{A+B}{2}))

= λ sin(\frac{A-B}{2}) (2cos(\frac{A+B}{2}) sin(\frac{A+B}{2}) )

By using trigonometric formula,

2 sin a cos a = sin 2a

= λ sin(\frac{A-B}{2}) (sin2(\frac{A+B}{2}) )

= λ sin(\frac{A-B}{2}) sin(A+B)

= λ sin(\frac{A-B}{2}) sin(π-C) (A+B+C = π)

= λ sin(\frac{A-B}{2}) sin(C)



= (λ sin C) sin(\frac{A-B}{2})

= c sin(\frac{A-B}{2})

As, LHS = RHS

Hence Proved!

Question 6: In △ABC, prove the following:

\frac{c}{a-b} = \frac{tan(\frac{A}{2})+tan(\frac{B}{2})}{tan(\frac{A}{2})-tan(\frac{B}{2})}

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS, we have

\frac{c}{a-b} = \frac{\lambda sinC}{\lambda(sinA-sinB)}

\frac{c}{a-b} = \frac{sinC}{sinA-sinB}



By using trigonometric identities,

sin A – sin B = 2 sin(\frac{A-B}{2}) cos(\frac{A+B}{2})

sin 2a = 2 sin a cos a

LHS = \frac{2sin\frac{C}{2}cos\frac{C}{2}}{2 sin\frac{A-B}{2}cos\frac{A+B}{2}}\\ = \frac{sin\frac{\pi-(A+B)}{2}cos\frac{C}{2}}{sin\frac{A-B}{2}cos\frac{A+B}{2}}\\ = \frac{sin(\frac{\pi}{2}-\frac{A+B}{2})cos\frac{C}{2}}{sin\frac{A-B}{2}cos\frac{A+B}{2}}\\ = \frac{cos(\frac{A+B}{2})cos\frac{C}{2}}{sin\frac{A-B}{2}cos\frac{A+B}{2}}\\ = \frac{cos\frac{C}{2}}{sin\frac{A-B}{2}} ……………………….(1)

Now considering RHS, we have

RHS = \frac{tan(\frac{A}{2})+tan(\frac{B}{2})}{tan(\frac{A}{2})-tan(\frac{B}{2})}\\ = \frac{\frac{sin(\frac{A}{2})}{cos(\frac{A}{2})}+\frac{sin(\frac{B}{2})}{cos(\frac{B}{2})}}{\frac{sin(\frac{A}{2})}{cos(\frac{A}{2})}-\frac{sin(\frac{B}{2})}{cos(\frac{B}{2})}}

Cross multiplying we get,

= \frac{\frac{sin(\frac{A}{2})cos(\frac{B}{2})+sin(\frac{B}{2})cos(\frac{A}{2})}{cos(\frac{A}{2})cos(\frac{B}{2})}}{\frac{sin(\frac{A}{2})cos(\frac{B}{2})-sin(\frac{B}{2})cos(\frac{A}{2})}{cos(\frac{A}{2})cos(\frac{B}{2})}}\\ = \frac{sin(\frac{A}{2})cos(\frac{B}{2})+sin(\frac{B}{2})cos(\frac{A}{2})}{sin(\frac{A}{2})cos(\frac{B}{2})-sin(\frac{B}{2})cos(\frac{A}{2})}

By using trigonometric identities,

sin a cos b + cos a sin b = sin (a+b)



sin a cos b – cos a sin b = sin (a-b)

= \frac{sin(\frac{A}{2}+\frac{B}{2})}{sin(\frac{A}{2}-\frac{B}{2})}\\ = \frac{sin(\frac{A+B}{2})}{sin(\frac{A-B}{2})}\\ = \frac{sin(\frac{\pi-C}{2})}{sin(\frac{A-B}{2})}\\ = \frac{sin(\frac{\pi}{2}-\frac{C}{2})}{sin(\frac{A-B}{2})}\\ = \frac{cos(\frac{C}{2})}{sin(\frac{A-B}{2})} ……………………….(2)

As, LHS = RHS

Hence Proved!

Question 7: In △ABC, prove the following:

\frac{c}{a+b} = \frac{1-tan(\frac{A}{2})tan(\frac{B}{2})}{1+tan(\frac{A}{2})tan(\frac{B}{2})}

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS, we have

\frac{c}{a+b} = \frac{\lambda sinC}{\lambda(sinA+sinB)}\\ \frac{c}{a+b} = \frac{sinC}{sinA+sinB}

By using trigonometric identities,



sin A + sin B = 2 sin(\frac{A+B}{2}) cos(\frac{A-B}{2})

sin 2a = 2 sin a cos a

LHS = \frac{2sin\frac{C}{2}cos\frac{C}{2}}{2 sin\frac{A+B}{2}cos\frac{A-B}{2}}\\ = \frac{sin(\frac{C}{2})cos(\frac{\pi-(A+B)}{2})}{sin\frac{A+B}{2}cos\frac{A-B}{2}}\\ = \frac{sin(\frac{C}{2})cos(\frac{\pi}{2}-\frac{A+B}{2})}{sin\frac{A+B}{2}cos\frac{A-B}{2}}\\ = \frac{sin(\frac{C}{2})sin(\frac{A+B}{2})}{sin\frac{A+B}{2}cos\frac{A-B}{2}}\\ = \frac{sin\frac{C}{2}}{cos(\frac{A-B}{2})} ………………………….(1)

Now considering RHS, we have

RHS = \frac{1- tan(\frac{A}{2})tan(\frac{B}{2})}{1+tan(\frac{A}{2})tan(\frac{B}{2})}\\ = \frac{1-\frac{sin(\frac{A}{2})}{cos(\frac{A}{2})}\frac{sin(\frac{B}{2})}{cos(\frac{B}{2})}}{1+\frac{sin(\frac{A}{2})}{cos(\frac{A}{2})}\frac{sin(\frac{B}{2})}{cos(\frac{B}{2})}}

Cross multiplying we get,

= \frac{\frac{cos(\frac{A}{2})cos(\frac{B}{2})-sin(\frac{A}{2})sin(\frac{B}{2})}{cos(\frac{A}{2})cos(\frac{B}{2})}}{\frac{cos(\frac{A}{2})cos(\frac{B}{2})+sin(\frac{A}{2})sin(\frac{B}{2})}{cos(\frac{A}{2})cos(\frac{B}{2})}}\\ = \frac{cos(\frac{A}{2})cos(\frac{B}{2})-sin(\frac{A}{2})sin(\frac{B}{2})}{cos(\frac{A}{2})cos(\frac{B}{2})+sin(\frac{A}{2})sin(\frac{B}{2})}

By using trigonometric identities,

cos a cos b + sin a sin b = cos (a-b)

cos a cos b – sin a sin b = cos (a+b)



= \frac{cos(\frac{A}{2}+\frac{B}{2})}{cos(\frac{A}{2}-\frac{B}{2})}\\ = \frac{cos(\frac{A+B}{2})}{cos(\frac{A-B}{2})}\\ = \frac{cos(\frac{\pi-C}{2})}{cos(\frac{A-B}{2})}\\ = \frac{cos(\frac{\pi}{2}-\frac{C}{2})}{cos(\frac{A-B}{2})}\\ = \frac{sin(\frac{C}{2})}{cos(\frac{A-B}{2})} ……………………….(2)

As, LHS = RHS

Hence, Proved!

Question 8: In △ABC, prove the following:

\frac{a+b}{c} = \frac{cos(\frac{A-B}{2})}{sin(\frac{C}{2})}

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS, we have

\frac{a+b}{c} = \frac{\lambda(sinA+sinB)}{\lambda sinC}\\ \frac{a+b}{c} = \frac{sinA+sinB}{sinC}

By using trigonometric identities,

sin A + sin B = 2 sin(\frac{A+B}{2}) cos(\frac{A-B}{2})



sin 2a = 2 sin a cos a

LHS = \frac{2 sin\frac{A+B}{2}cos\frac{A-B}{2}}{2sin\frac{C}{2}cos\frac{C}{2}}\\ = \frac{sin\frac{A+B}{2}cos\frac{A-B}{2}}{sin(\frac{C}{2})cos(\frac{\pi-(A+B)}{2})}\\ = \frac{sin\frac{A+B}{2}cos\frac{A-B}{2}}{sin(\frac{C}{2})cos(\frac{\pi}{2}-\frac{A+B}{2})}\\ = \frac{sin\frac{A+B}{2}cos\frac{A-B}{2}}{sin(\frac{C}{2})sin(\frac{A+B}{2})}\\ = \frac{cos(\frac{A-B}{2})}{sin\frac{C}{2}}

As, LHS = RHS

Hence, Proved!

Question 9: In △ABC, prove the following:

sin(\frac{B-C}{2}) = (\frac{b-c}{a})cos(\frac{A}{2})

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering RHS, we have

(\frac{b-c}{a})cos(\frac{A}{2}) = (\frac{\lambda(sinB-sinC)}{\lambda sinA})cos(\frac{A}{2})\\ = (\frac{sinB-sinC}{sinA})cos(\frac{\pi-(B+C)}{2})

By using trigonometric identities,

sin A – sin B = 2 sin(\frac{A-B}{2}) cos(\frac{A+B}{2})

LHS = (\frac{2 sin\frac{B-C}{2}cos\frac{B+C}{2}}{sinA})cos(\frac{\pi}{2}-\frac{B+C}{2})\\ = (\frac{2 sin\frac{B-C}{2}cos\frac{B+C}{2}}{sinA})sin(\frac{B+C}{2})\\ = (\frac{[2 sin(\frac{B+C}{2}) cos\frac{B+C}{2}] sin\frac{B-C}{2}}{sinA})

By using trigonometric identities,

2 sin a cos a = sin 2a

= (\frac{sin(2(\frac{B+C}{2}))sin(\frac{B-C}{2})}{sinA})\\ = (\frac{sin(B+C)sin(\frac{B-C}{2})}{sinA})\\ = (\frac{sin(\pi-(B+C))sin(\frac{B-C}{2})}{sinA})\\ = (\frac{sin(A)sin(\frac{B-C}{2})}{sinA})\\ = sin(\frac{B-C}{2})

As, LHS = RHS

Hence, Proved!

Question 10: In △ABC, prove the following:

\frac{a^2-c^2}{b^2} = \frac{sin(A-C)}{sin(A+B)}

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda



Considering LHS, we have

\frac{a^2-c^2}{b^2} = \frac{[\lambda sinA]^2-[\lambda sinC]^2}{[\lambda sinB]^2}\\ = \frac{(\lambda)^2 [sin^2A - sin^2C]}{(\lambda)^2[sin^2B]}\\ = \frac{sin^2A - sin^2C}{sin^2B}

By using trigonometric identities,

sin2 a – sin2 b = sin(a+b) sin (a-b)

LHS = \frac{sin(A+C) sin(A-C)}{sin^2B}\\ = \frac{sin(A+C) sin(A-C)}{sin^2(\pi-(A+C))}\\ = \frac{sin(A+C) sin(A-C)}{sin^2(A+C)}\\ = \frac{sin(A-C)}{sin(A+C)}\\

As, LHS = RHS

Hence, Proved!

Question 11: In △ABC, prove the following:

b sin B – c sin C = a sin (B-C)

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

And, cosine rule,

cos \hspace{0.1cm} a = \frac{b^2+c^2-a^2}{2bc}

Considering RHS, we have

RHS = a sin (B-C)

By using trigonometric identities,

sin(a-b) = sin a cos b – cos a sin b

= a (sin B cos C – cos B sin C)

= a ((bλ)(\frac{a^2+b^2-c^2}{2ab}) (\frac{a^2+c^2-b^2}{2ac}) (cλ))

= λ(\frac{a^2+b^2-c^2}{2} - \frac{a^2+c^2-b^2}{2})

= 2λ(\frac{b^2-c^2}{2})



= λb2-λc2

= b(λb) – c(λc)

= b(sin B) – c(cos C)

As, LHS = RHS

Hence, Proved!

Question 12: In △ABC, prove the following:

a2 sin (B-C) = (b2-c2) sin A

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering RHS, we have

RHS = (b2-c2) sin A

= λ2 sin A(sin2 B – sin2 C)

By using trigonometric identities,

sin2 a – sin2 b = sin(a+b) sin (a-b)

= λ2 sin A(sin(B+C) sin (B-C))

= λ2 sin A(sin(\pi-A) sin (B-C))

= λ2 sin A(sin(A) sin (B-C))

= λ2 sin2 A sin (B-C)

= (λ sin A)2 sin (B-C)

= a2 sin (B-C)

As, LHS = RHS

Hence, Proved!

Question 13: In △ABC, prove the following:

\frac{\sqrt{sin A}-\sqrt{sin B}}{\sqrt{sin A}+\sqrt{sin B}} = \frac{a+b-2\sqrt{ab}}{a-b}

Solution:

Considering LHS, we have

LHS =\frac{\sqrt{sin A}-\sqrt{sin B}}{\sqrt{sin A}+\sqrt{sin B}}

Rationalizing the denominator, we get

= \frac{\sqrt{sin A}-\sqrt{sin B}}{\sqrt{sin A}+\sqrt{sin B}} \times \frac{\sqrt{sin A}-\sqrt{sin B}}{\sqrt{sin A}-\sqrt{sin B}}\\ = \frac{(\sqrt{sin A}-\sqrt{sin B})^2}{(\sqrt{sin A})^2-(\sqrt{sin B})^2}\\ = \frac{(\sqrt{sin A})^2+(\sqrt{sin B})^2-2(\sqrt{sin A})(\sqrt{sin B})}{sin A-sin B}\\ = \frac{sin A+sin B-2\sqrt{sin A sin B}}{sin A-sin B}

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

= \frac{\frac{a}{\lambda}+\frac{b}{\lambda}-2\sqrt{\frac{a}{\lambda} \frac{b}{\lambda}}}{\frac{a}{\lambda}-\frac{b}{\lambda}}\\ = \frac{\frac{1}{\lambda}(a+b-2\sqrt{ab})}{\frac{1}{\lambda}(a-b)}\\ = \frac{a+b-2\sqrt{ab}}{a-b}

As, LHS = RHS



Hence, Proved!

Question 14: In △ABC, prove the following:

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

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS, we have

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

= λ sin A (sin B – sin C) + λ sin B (sin C – sin A) + λ sin C (sin A – sin B)

= λ sin A sin B – λ sin A sin C + λ sin B sin C – λ sin B sin A) + λ sin C sin A – λ sin C sin B

= 0

As, LHS = RHS

Hence, Proved!

Question 15: In △ABC, prove the following:

\frac{a^2 sin (B-C)}{sin A} + \frac{b^2 sin (C-A)}{sin B} + \frac{c^2 sin (A-B)}{sin C} = 0

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS, we have

LHS =\frac{a^2 sin (B-C)}{sin A} + \frac{b^2 sin (C-A)}{sin B} + \frac{c^2 sin (A-B)}{sin C}

=\frac{(\lambda sin A) sin (B-C)}{sin A} + \frac{(\lambda sin B) sin (C-A)}{sin B} + \frac{(\lambda sin C) sin (A-B)}{sin C}

= λ2 sin A sin (B-C) + λ2 sin B sin (C-A) + λ2 sin C sin (A-B)

By using trigonometric identities,

sin(a-b) = sin a cos b – cos a sin b

= λ2 (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])

= λ2 (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)

= λ2 (0)

= 0

As, LHS = RHS

Hence, Proved!




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