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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.
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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