LetABCbe a triangle such thatβ ACB=Ο6and leta, bandcdenote the lengths of the sides opposite toA, BandCrespectively. The value(s) of x for whicha=x2+x+1, b=x2π΅andc=2x+1is (are)β(2+β3)1+β32+β34β3LetABCbe a triangle such thatβ ACB=Ο6and leta, bandcdenote the lengths of the sides opposite toA, BandCrespectively. The value(s) of x for whicha=x2+x+1, b=x2π΅andc=2x+1is (are)β(2+β3)1+β32+β34β3ABCABCABCAABBCCβ ACB=Ο6β ACB=Ο6β ACB=Ο6β β AACCBB==Ο6Ο6Ο6Ο6ΟΟΟ666a, ba, ba, baa,, bbcccccA, BA, BA, BAA,, BBCCCCCa=x2+x+1, b=x2π΅a=x2+x+1, b=x2π΅a=x2+x+1, b=x2π΅aa==x2xxx22222++xx++11,, bb==x2xxx22222βπ΅1c=2x+1c=2x+1c=2x+1cc==22xx++11β(2+β3)β(2+β3)β(2+β3)β(2+β3)ββ((22++β3β3ββ3333))1+β31+β31+β31+β311++β3β3ββ33332+β32+β32+β32+β322++β3β3ββ33334β34β34β34β344β3β3ββ3333A1+β31+β31+β31+β31+β311++β3β3ββ3333Bβ(2+β3)β(2+β3)β(2+β3)β(2+β3)β(2+β3)ββ((22++β3β3ββ3333))C4β34β34β34β34β344β3β3ββ3333D2+β32+β32+β32+β32+β322++β3β3ββ3333?
1+β3
β(2+β3)
4β3
2+β3
Solution:
The correct option is B1+β3Using cosine rule forβ Ccosβ C=a2+b2βc22ab