# Class 11 RD Sharma Solutions – Chapter 10 Sine and Cosine Formulae and Their Applications – Exercise 10.1 | Set 2

### a2 (cos2 B – cos2 C) + b(cos2 C – cos2 A) + c(cos2 A – cos2 B) = 0

Solution:

According to the sine rule

Considering LHS, we have

= a2 (cos2 B – cos2 C) + b(cos2 C – cos2 A) + c(cos2 A – cos2 B)

By using trigonometric formula,

cos2 a = 1 – sin2 a

= Î»2 sin2 A(1-sin2 B – (1-sin2 C)) + Î»2 sin2 B(1-sin2 C – (1-sin2 A)) + Î»2 sin2 C(1-sin2 A – (1-sin2 B))

= Î»2 sin2 A(1-sin2 B – 1+ sin2 C) + Î»2 sin2 B(1-sin2 C – 1 + sin2 A) + Î»2 sin2 C(1-sin2 A – 1+sin2 B)

= Î»2 sin2 A(sin2 C – sin2 B) + Î»2 sin2 B(sin2 A – sin2 C) + Î»2 sin2 C(sin2 B – sin2 A)

= Î»2 (sin2 A sin2 C – sin2 A sin2 B + sin2 B sin2 A – sin2 B sin2 C + sin2 C sin2 B – sin2 C sin2 A)

= Î»2 (0)

=0

As LHS = RHS

Hence, proved!!

### b cos B + c cos C = a cos (B-C)

Solution:

According to the sine rule

Considering LHS, we have

= b cos B + c cos C

= Î» sin B cos B + Î» sin C cos C

= Î» (sin B cos B + sin C cos C)

=(2 sin B cos B + 2 sin C cos C)

By using trigonometric formula,

2 sin a cos a = sin 2a

=(sin 2B + sin 2C) ………………………..(1)

Now considering RHS, we have

= a cos (B-C)

= Î» sin A cos (B-C)

=(2 sin A cos (B-C))

By using trigonometric formula,

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

=(sin 2C + sin2B) ……………………….(2)

As LHS = RHS

Hence, proved!!

### Question 18: In â–³ABC, prove the following:

Solution:

According to the sine rule

Considering LHS, we have

By using trigonometric formula,

cos 2a = 1-2sin2a

As LHS = RHS

Hence, proved!!

### Question 19: In â–³ABC, prove the following:

Solution:

According to the sine rule

Considering LHS, we have

Now taking,

By using trigonometric formula,

cos a + cos b = 2 coscos

cos a – cos b = -2 sinsin

sin a + sin b = 2 sincos

By using trigonometric formula,

2 cossin= sin a – sin b

………………(1)

Similarly, we can prove,

……………….(2)

………………..(3)

Adding (1), (2) and (3), we get

= 0

As LHS = RHS

Hence, proved!!

### Question 20: In â–³ABC, prove the following:

Solution:

According to the sine rule

Considering LHS, we have

Now taking,

=a cossin

=(sin B – sin C)

=(a sin B – a sin C)

=(a sin B – a sin C)

=(b sin A – a sin C) ………………..(1)

Similarly, we can prove,

bsin=(b sin C – b sin A) ……………….(2)

csin=(a sin C – b sin C) ………………..(3)

Adding (1), (2) and (3), we get

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

=(b sin A – a sin C + b sin C – b sin A + a sin C – b sin C)

= 0

As LHS = RHS

Hence, proved!!

### Question 21: In â–³ABC, prove the following:

Solution:

According to the sine rule

Considering equation, we have

Now taking,

Similarly, we can prove,

From, (1), (2) and (3), we get

Hence, proved!!

### a cos A + b cos B + c cos C = 2b sin A sin C = 2c sin A sin B

Solution:

According to the sine rule

Considering LHS equation, we have

a cos A + b cos B + c cos C = Î» sin A cos A + Î» sin B cos B + Î» sin C cos C

= Î» (sin A cos A + sin B cos B + sin C cos C)

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

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

By using trigonometric formula,

2 sin a cos a = sin 2a

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

By using trigonometric formula,

sin a + sin b = 2 sincos

=+ 2 sin C cos C)

=(2 sin (A+B) cos(A-B) + 2 sin C cos C)

=(2 sin (Ï€-C) cos(A-B) + 2 sin C cos C)

=(2 sin C cos(A-B) + 2 sin C cos C)

=(cos(A-B) + cos C)

= Î» sin C (cos(A-B) + cos (Ï€-(A+B)))

= Î» sin C (cos(A-B) + (-cos (A+B)))

= Î» sin C (cos(A-B) – cos (A+B))

= Î» sin C (2 sin A sin B)

= 2 Î» sin A sin B sin C

Now putting Î» sin C = c and Î» sin B = b, we get

2 c sin A sin B and 2 b sin A sin C

Hence, proved!!

### a (cos B cos C + cos A) = b (cos C cos A + cos B) = c (cos C cos A + cos C)

Solution:

According to the sine rule

Considering equation, we have

a (cos B cos C + cos A) = Î» sin A (cos B cos C + cos A)

= Î» (sin A cos B cos C + sin A cos A)

= Î» ((2 sin A cos B) + sin A cos A)

= Î» ((sin (A+B) + sin (A-B)) + sin A cos A)

= Î» ((cos C sin (A+B) + cos C sin (A-B)) + sin A cos A)

= Î» ((\frac{1}{2} (2 cos C sin (A+B) + 2 cos C sin (A-B))) + sin A cos A)

= Î» ((2 cos C sin (A+B) + 2 cos C sin (A-B)) + sin A cos A)

By using trigonometric formula,

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

= Î» ((sin (A+B+C) + sin (A+B-C) + sin (A-B+C) + sin (A-B-C)) + sin A cos A)

= Î» ((sin (Ï€) + sin ((Ï€-C)-C) + sin ((Ï€-B)-B) + sin (A-(B+C)) +)

= Î» ((0 + sin (Ï€-2C) + sin (Ï€-2B) + sin (2A-Ï€) +)

= Î» ((sin 2C + sin 2B – sin 2A +)

=(sin 2C + sin 2B + sin 2A)

Similarly,

b (cos C cos A + cos B) =(sin 2C + sin 2B + sin 2A)

c (cos C cos A + cos C) =(sin 2C + sin 2B + sin 2A)

Hence, proved!!

### a (cos C – cos B) = 2 (b-c)

Solution:

According to the sine rule

Considering equation, we have

a (cos C – cos B) = Î» sin A (cos C – cos B)

= Î» (sin A cos C – sin A cos B)

=(2 sin A cos C – 2 sin A cos B)

By using trigonometric formula,

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

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

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

=(sin (Ï€-B) + sin (A-C) – sin (Ï€-C) – sin (A-B))

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

By using trigonometric formula,

sin a – sin b = 2 sincos

= Î»

= Î»

= Î» sin

By using trigonometric formula,

sin a + sin b = 2 sincos

= Î» sin

= Î» sin

= Î» sin(2 sin A cos)

= 2 Î» sin(2 sincos) cos

= 4 Î» sinsin[Tex]cos^2(\frac{A}{2})[/Tex]

= 4 Î» sincos[Tex]cos^2(\frac{A}{2})[/Tex]

= 4 Î» sincos[Tex]cos^2(\frac{A}{2})[/Tex]

= 4 Î» sincos[Tex]cos^2(\frac{A}{2})[/Tex]

= 2 Î» (2 sincos)

By using trigonometric formula,

2 cossin= sin a – sin b

= 2 Î» (sin B – sin A)

= 2 (Î» sin B – Î» sin A)

= 2 (b – a)

As LHS = RHS

Hence, proved!!

### b cos Î¸ = c cos(A-Î¸)+a cos(C+Î¸)

Solution:

According to the sine rule

Considering RHS, equation, we have

c cos(A-Î¸)+a cos(C+Î¸) = Î» sin C cos(A-Î¸) + Î» sin A cos(C+Î¸)

= Î» (sin C cos(A-Î¸) + sin A cos(C+Î¸))

=

By using trigonometric formula,

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

By using trigonometric formula,

sin (a+b) + sin (a-b) = 2 sin a cos b

=

= Î» sin B cos Î¸

= b cos Î¸

As LHS = RHS

Hence, proved!!

### Question 26: In â–³ABC, if sin2A + sin2B = sin2C. Show that the triangle is right-angled.

Solution:

According to the sine rule

Considering equation, we have

sin2 A + sin2 B = sin2 C

a2 + b2 = c2

Hence, proved, the triangle is right-angled as c as hypotenuse.

### Question 27: In â–³ABC, if a2, b2, and c2 are in AP. Prove that cot A, cot B, and cot C are also in AP.

Solution:

We have a2, b2 and c2 in AP

2a2, 2b2 and 2c2 are also in AP

(a2+b2+c2)-2a2, (a2+b2+c2)-2b2 and (a2+b2+c2)-2c2 are also in AP

b2+c2-a2, a2+c2-b2 and a2+b2-c2 are also in AP

,andare also in AP

,andare also in AP

According to the cosine rule

,andare also in AP

,andare also in AP

According to the sine rule

,andare also in AP

cot A, cot B and cot C are also in AP

Hence, proved !!

### Question 28: The upper part of a tree broken by the wind makes an angle of 30Â° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. Using sine rule, find the height of the tree.

Solution:

Suppose BD be the tree and the upper part of the tree is broken over by the wind at point A.

The total height of the tree is x+y.

In â–³ABC, âˆ C = 30Â° and âˆ B = 90Â°

âˆ A = 60Â° (Triangle angle sum property)

According to the sine rule

2x == y

Hence, x == 5âˆš3

and y == 10âˆš3

So, the height of tree x+y = 5âˆš3+10âˆš3

= 15âˆš3 m

### Question 29: At the foot of a mountain, the elevation of it summit is 45Â°, after ascending 1000m towards the mountain up a slope of 30Â° inclination, the elevation is found to be 60Â°. Find the height of the mountain.

Solution:

Suppose, AB is a mountain of height t+x.

c = 1000m

In â–³DFC,

sin 30Â° =

x == 500 m

And, tan 30Â° =

y = 500âˆš3 m

Now, In â–³ABC

tan 45Â° =

1 =

500âˆš3+z = t+500

500(âˆš3-1)+z = t …………………..(1)

tan 60Â° =

âˆš3 =

t = âˆš3z ………………….(2)

From (1) and (2), we get

z = 500 m

t = 500âˆš3 m

So, the height of the mountain = t+x = 500âˆš3 + 500 = 500(âˆš3+1)m

### Question 30: A person observes the angle of elevation of the peak of a hill from a station to be Î±. He walks c metres along a slope inclined at an angle Î² and finds the angle of elevation of the peak of the hill to be Î³. Show that the height of the peak above the ground is

Solution:

Suppose, AB is a peak whose height above the ground is t+x,

In â–³DFC,

sin Î² =

x = c sin Î² ………………………..(1)

And, tan Î² =

= c cos Î² ………………………..(2)

tan Î³ =

z = t cot Î³ ………………………(3)

Now, In â–³ABC

tan Î±=

t +x = tan Î± (y+z)

From (1), (2) and (3), we get

t + c sin Î² = tan Î± (c cos Î²+t cot Î³)

t + c sin Î² = c tan Î± cos Î² + t tan Î± cot Î³

t – t tan Î± cot Î³ = c tan Î± cos Î² – c sin Î²

t(1 – tan Î± cot Î³) = c (tan Î± cos Î² – sin Î²)

Now,

Hence proved !!

### Question 31: If the sides a, b and c of â–³ABC are in H.P. Prove thatandare in H.P

Solution:

If the sides a, b and c of â–³ABC are in H.P

Then,,andare in AP

According to the sine rule

By using trigonometric formula,

sin a – sin b = 2 sincos

Divide by, we get

Hence,andare in H.P

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