A) \[\angle B={{60}^{o}}\]
B) \[\angle B={{30}^{o}}\]
C) \[\angle C={{60}^{o}}\]
D) \[\angle A+\angle C={{90}^{o}}\]
Correct Answer: A
Solution :
\[{{(a+c)}^{2}}-{{b}^{2}}=3ac\Rightarrow {{a}^{2}}+{{c}^{2}}-{{b}^{2}}=ac\] But\[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}=\frac{1}{2}\Rightarrow B=\frac{\pi }{3}\].
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