Question

please help please please​

Answer 1

Answer:

Step-by-step explanation:

Consider any triangle with the sides as a,b,c and A,B,C as the angles. the sine Formula is a/SinA = b/SinB = c/SinC

And sum of the angles of a triangle = 180°

it means A + B + C = 180,

A = 180 - (B + C),

B + C = 180 - A

now, we have to prove,

(b+c) Cos (B+C/2) = a Cos(B-C/2)

we can also write this as

=> b+c/a = Cos(B-C/2) / Cos(B+C/2)

now, From the sine Formula,

a/SinA = b/SinB = c/SinC = k

a = kSinA, b = kSinB, c = kSinC

So, L.H.S,

b+c/a = kSinB + kSinC/ kSinA

Remember Sinx + Siny = 2Sin(x+y/2)Cos(x-y/2) and Sin2A = 2SinACosA.

          = 2kSin(B+C/2)Cos(B-C/2)/2kSinA/2CosA/2

          = SIn((180-A)/2)Cos(B-C/2)/Sin((180 - (B+C))/2)CosA/2

          = Sin(90-A/2)Cos(B-C/2) / Sin(90 - B+C/2)CosA/2

Remember Sin (90 - θ) = Cosθ

          = Cos(A/2)Cos(B-C/2)/ Cos(B+C/2).CosA/2

          = Cos(B-C/2)/Cos(B+C/2)

          = R.H.S.

Hence proved.