let 1< m<3. In a triangle ABC if 2b = (m+1)a andcos A = [(m-1)(m+3)/4m]1/2 . prove that there r 2 values for the third side one of which is m times the other.

Dear

cos A = [(m-1)(m+3)/4m]^1/2

calculate sinA = [1-cos²A]^1/2

                    = [(m+1)(3-m)/4m]^1/2

now sinB /b =sinA /a

sinB = (m+1)/2 sinA

         =  [(m+1)³(3-m)/16m]^1/2

now calculate CosB = ± [1-sin²B]^1/2

                                    =± [(m-1)³(m+3)/16m]^1/2

now sinC =sin(180 -A-B)

                 =sin(A+B)

                 =sinA cosB + cosA sinB

                 =[(m-1)(m+3)(1+m)(3-m)/16m²]^1/2 [(1+m) ±(m-1)]

for + value   SinC =[(m-1)(m+3)(1+m)(3-m)/16m²]^1/2 [2m]

for - alue     sinC =[(m-1)(m+3)(1+m)(3-m)/16m²]^1/2 [(2]

clearly one value is m times the other

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