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Class 11 RD Sharma Solutions – Chapter 10 Sine and Cosine Formulae and Their Applications – Exercise 10.2 | Set 2
  • Last Updated : 30 Apr, 2021

Question 11. In any ∆ABC, prove the following: a cos A + b cos B + c cos C = 2b sin A sin C.

Solution:

According to sine rule in ΔABC,

a/sin A = b/sin B = c/sin C = k (constant) 

L.H.S. = a cos A + b cos B + c cos C

= k sin A cos A + k sin B cos B + k sin C cos C



= (k/2) [2 sin A cos A + 2 sin B cos B + 2 sin C cos C]

= (k/2) [sin 2A + sin 2B + sin 2C]

= (k/2) [sin 2A + sin 2B + sin 2C]

= (k/2) [sin(A+B) cos(A–B) + sin C cos C]

= (k/2) [sin(π–C) cos(A–B) + sin C cos(π–(A+B))]

= (k/2) [sin C cos(A–B) + sin C cos(A+B)]

= k sin C [2 sin A sin B]

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

= 2b sin A sin C

= R.H.S.

Hence, proved.

Question 12. Prove that a2 = (b+c)2 – 4bc cos2A/2

Solution:

According to Cosine rule,

=> cos A = (b2 + c2 – a2)/2bc

=> 2bc cos A = b2 + c2 – a2

=> a2 = b2 + c2 – 2bc cos A

=> a2 = b2 + c2 – 2bc (2cos2A/2 – 1)

=> a2 = b2 + c2 – 4bc cos2A/2 + 2bc



=> a2 = (b+c)2 – 4bc cos2A/2

Hence, proved.

Question 13. Prove that 4(bc cos2A/2 + ac cos2B/2 + ab cos2C/2) = (a + b + c)2 

Solution:

According to Cosine rule,

cos A = (b2 + c2 – a2)/2bc  . . . . (1)

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

cos C = (a2 + b2 – c2)/2ab  . . . . (3)

We have, 

L.H.S. = 4(bc cos2A/2 + ac cos2B/2 + ab cos2C/2)

= 2(2bc cos2A/2 + 2ac cos2B/2 + 2ab cos2C/2)



= 2[bc(1–cos A) + ac(1–cos B) + ab(1–cos C)]

= 2bc – 2bc cos A + 2ac – 2ac cos B + 2ab – 2ab cos C

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

L.H.S. = 2bc – [b2 + c2 – a2] + 2ac – [a2 + c2 – b2] + 2ab – [a2 + b2 – c2]

= 2bc – b2 – c2 + a2 + 2ac – a2 – c2 + b2 + 2ab – a2 – b2 + c2

= a2 + b2 + c2 + 2ab + 2bc + 2ca

= (a + b + c)2 

= R.H.S.

Hence, proved.

Question 14. In a ΔABC, prove that sin3A cos(B–C) + sin3B cos(C–A)+ sin3C cos(A–B) = 3 sinA sinB sin C

Solution:

We have, 

L.H.S. = sin3A cos(B–C) + sin3B cos(C–A)+ sin3C cos(A–B)

= sin2A sin A cos(B–C) + sin2B sin B cos(C–A)+ sin2C sin C cos(A–B)

= sin2A sin(π–(B+C)) cos(B–C) + sin2B sin(π–(A+C)) cos(C–A)+ sin2C sin(π–(A+B)) cos(A–B)

= sin2A sin(B+C) cos(B–C) + sin2B sin(A+C) cos(C–A)+ sin2C sin(A+B) cos(A–B)

= sin2A [sin2B + sin2C] + sin2B [sin2A + sin2C]+ sin2C [sin2A + sin2B]

= sin2A [2sinB cosB + 2sinC cosC] + sin2B [2sinA cosA+ 2sinC cosC]+ sin2C [2sinA cosA + 2sinB cosB]

= 2sin2A sinB cosB + 2sin2A sinC cosC + 2sin2B sinA cosA + 2sin2B sinC cosC + 2sin2C sinA cosA + 2sin2C sinB cosB

According to sine rule in ΔABC,

sin A/a = sin B/b = sin C/c = k (constant) 

So, L.H.S. becomes,

= 2a2k2 (bk) (cosB) + 2a2k2 (ck) (cosC) + 2b2k2 (ak) (cosA) + 2b2k2 (ck) (cosC) + 2c2k2 (ak) (cosA) + 2c2k2 (bk) cosB

= abk3 (a cosB + b cosA) + ack3 (a cosC + c cosA) + bck3 (c cosB + b cosC)

= abck3 + abck3 + abck3

= 3abck3

= 3 (ak) (bk) (ck)

= 3 sin A sin B sin C

= R.H.S.

Hence, proved.

Question 15. In any ΔABC, (b+c)/12 = (c+a)/13 = (a+b)/15, prove that (cosA)/2 = (cosB)/7 = (cosC)/11.

Solution:

We are given,

(b+c)/12 = (c+a)/13 = (a+b)/15 = k (say)

=> b + c = 12k  . . . . (1)

=> c + a = 13k  . . . . (2)

=> a + b = 15k  . . . . (3)

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

=> b + c + c + a + a + b = 12k + 13k + 15k

=> 2a + 2b + 2c = 40k

=> a + b + c = 20k  . . . . (4)

From (1), (2), (3) and (4), we get,

=> a = 8k, b = 7k and c = 5k

According to Cosine formula,

cos A = (b2 + c2 – a2)/2bc

= (49k2 + 25k2 – 64k2)/70k2

= 10/70

= 1/7

cos B = (a2 + c2 – b2)/2ac

= (64k2 + 25k2 – 49k2)/80k2

= 40/80

= 1/2

cos C = (a2 + b2 – c2)/2ab

= (64k2 + 49k2 – 25k2)/112k2

= 88/112

= 11/14

Hence, cosA : cosB : cosC = (1/7) : (1/2) : (11/14) = 2 : 7 : 11

=> (cosA)/2 = (cosB)/7 = (cosC)/11

Hence, proved.

Question 16. In ΔABC, if ∠B = 60o, prove that (a + b + c) (a – b + c) = 3ca.

Solution:

We have, ∠B = 60o

According to Cosine formula,

cosB = (a2 + c2 – b2)/2ac

=> cos 60o = (a2 + c2 – b2)/2ac 

=> 1/2 = (a2 + c2 – b2)/2ac 

=> a2 + c2 – b2 = ac  . . . . (1)

Now L.H.S. = (a + b + c) (a – b + c)

= a2 – ab + ac + ab – b2 + bc + ca – bc + c2

= a2 + c2 – b2 + 2ac

=  ac + 2ac 

= 3ac 

= R.H.S.

Hence, proved.

Question 17. If in a ΔABC, cos2A + cos2B + cos2C = 1, prove that the triangle is right angled.

Solution:

We are given,

=> cos2A + cos2B + cos2C = 1

=> cos2A + cos2B = 1 – cos2C

=> cos2A + cos2B = sin2C

=> cos2A = sin2C − cos2B

=> −cos(B+C) cos(B−C) = cos2A

=> −cos(π−A) cos(B−C) = cos2A

=> cos2A − cosA cos(B−C) = 0

=> cosA (cosA − cos(B−C)) = 0

=> cosA [cos(π−(B+C)) − cos(B−C)] = 0

=> cosA [−cos(B+C) − cos(B−C)] = 0

=> cosA [−2 cosB cosC] = 0

=> cosA cos B cos C = 0

=> cosA = 0 or cosB = 0 or cosC = 0

=> A = 90o or B = 90o or C = 90o

Therefore, the triangle is right-angled.

Question 18. In a ΔABC, if cosC = sinA/2sinB, prove that the triangle is isosceles. 

Solution:

Here, we are given

=> cosC = sinA/2sinB

=> 2 sinB cosC = sinA

=> 2 (kb) [(a2+b2−c2)/2ab] = ka

=> a2 + b2 − c2 = a2

=> b2 = c2

=> b = c

Therefore, the triangle is isosceles.

Question 19. Two ships leave a port at the same time. One goes 24 km/hr in the direction N38oE and other travels 32 km/hr in the direction S52oE. Find the distance between the ships at the end of 3 hours.

Solution:

Let A be the point from where ships leave. AB is the distance travelled by one and AC is the distance travelled by the second in 3 hours. 

We have to find BC, the distance between the ships at the end of 3 hours. Here,

AB = 3(24) = 72 km and AC = 3(32) = 96 km.

Using Cosine formula in ΔABC, we get,

BC2 = AB2 + AC2 + 2 (AB) (AC) cos900

BC2 = 722 + 962

BC2 = 14400

BC = 120 km

Therefore, the distance between the ships at the end of 3 hours is 120 km.

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