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(B) We know that for any triangle $ABC$,the maximum value of $\cos A + \cos B + \cos C$ is $\frac{3}{2}$,which occurs if and only if $A = B = C = 60^{

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Equilateral

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(B) We know that for any triangle $ABC$,the maximum value of $\cos A + \cos B + \cos C$ is $\frac{3}{2}$,which occurs if and only if $A = B = C = 60^{\circ}$.Alternatively,using the identity $\cos A + \cos B + \cos C = 1 + 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{3}{2}$,we get $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{1}{8}$.Since the maximum value of $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$ is $\frac{1}{8}$ (by Jensen's inequality for concave functions),this equality holds only when $\frac{A}{2} = \frac{B}{2} = \frac{C}{2} = 30^{\circ}$,i.e.,$A = B = C = 60^{\circ}$.Thus,the triangle is equilateral.

If in a triangle $ABC$,$\cos A + \cos B + \cos C = \frac{3}{2}$,then the triangle is

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