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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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