In a triangle ABC, AB=AC and the length of the median from B to the side AC is 1. Find the cosine of the angle A for which the area of triangle is minimun.

Dear harshit

cosA =((b/2)² +c² -1) /bc

   b=c

 so   cos A = 5/4   -1/b² 

  b²=4/(5-4cosA)

nor area of triangle is =1/2 bc sinA

                              =1/2 b² sinA

                              =2sinA/(5-4cosA)

now differntiate and value of cos A for minimum value of area

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