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let 1< m<3. In a triangle ABC if 2b = (m+1)a and

cos 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.

6 years ago


Answers : (1)



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)


                 =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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