In a $$\triangle A B C, 2(b c \cos A+a c \cos B+a b \cos C)=$$

In a $$\triangle A B C, \frac{a}{b}=2+\sqrt{3}$$ and $$\angle C=60^{\circ}$$. Then, the measure of $$\angle A$$ is

If $$a=2, b=3, c=4$$ in a $$\triangle A B C$$, then $$\cos C=$$

In a $$\triangle A B C$$ $$(b+c) \cos A+(c+a) \cos B+(a+b) \cos C=$$