Question

a cosB-C/2= (b+c) cosB+C/2​

Answer 1

Answer:

Step-by-step explanation:

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

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

we know the sine rule of triangle i.e.

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

=> a = kSinA, b = kSinB, c = kSinC.

Consider L.H.S,

b + c/a = kSinB + kSinC / kSinA

= SinB + SinC / SinA

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

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we know that sum of all angles in triangle is 180°

=> A + B + C = 180°

=> B+C = 180 - A (Divide by 2 on both sides)

=> B+C/2 = 90 - A/2  ------ (1)

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Substitute the (1) above in place of B+C/2,

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

= CosA/2Cos(B-C/2) / SinA/2CosA/2

= Cos(B-C/2) / SinA/2

//Substitute the (1) above in place of SinA/2,

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

= Cos(B-C/2)/Cos(B+C/2)

= R.H.S