A) right angled
B) isosceles
C) acute
D) equilateral
Correct Answer: D
Solution :
\[\cos A+\cos B+\cos C=3/2\](given) \[\Rightarrow \] \[\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}+\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] \[+\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=3/2\] \[\Rightarrow \]\[(a+b-c){{(a-b)}^{2}}+(b+c-a){{(b-c)}^{2}}\] \[+(c+a-b){{(c-a)}^{2}}=0\] As we know, \[a+b>c,b+c>a,c+a>b\] \[\Rightarrow \]\[a=b=c\Rightarrow \]Triangle is an equilateral.
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