In a triangle ABC, if #a^4 -2(b^2+c^2)a^2 + b^4 + b^2c^2 + c^4 = 0#, then angle #A# is #60^o# or #120^o#?

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Answer 1 · 2017-12-13T08:49:48

Given

#a^4 -2(b^2+c^2)a^2 + b^4 + b^2c^2 + c^4 = 0#

#=>a^4 -2a^2b^2-2c^2a^2 + b^4 + 2b^2c^2 + c^4 = b^2c^2#

#=>(a^2 -b^2-c^2)^2 = b^2c^2#

#=>a^2 -b^2-c^2 =pm bc#

#=>b^2+c^2-a^2 =pm bc#

#=>(b^2+c^2-a^2)/(2bc) =pm1/2#

#=>cosA =pm1/2#

When

#cosA =1/2=cos60^@=>A=60^@#

When

#cosA =-1/2=cos120^@=>A=120^@#