A) Isosceles
B) Equilateral
C) Right angled
D) None of these
Correct Answer: B
Solution :
Putting \[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]in given relation, we get \[\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}+\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ac}\]\[+\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=\frac{3}{2}\] Þ \[a({{b}^{2}}+{{c}^{2}})+b({{c}^{2}}+{{a}^{2}})+c({{a}^{2}}+{{b}^{2}})\] \[={{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3abc\] Þ \[{{(b-c)}^{2}}(b+c-a)+{{(c-a)}^{2}}(c+a-b)\] \[+{{(a-b)}^{2}}(a+b-c)=0\] ?..(i) In triangle, \[b+c-a>0\]etc. and hence (i) will hold good if each factor is zero so that \[a=b=c\].
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