In a $\mathrm{△}ABC,\mathrm{\Sigma }(b+c)\cos A=$ 

Solution:

Using projection formulae, we have  

$\sum (b+c)\cos A$

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

$\begin{array}{l}=(b\cos C+c\cos B)+(c\cos A+a\cos C)+(a\cos B+b\cos A)\\ =a+b+c\end{array}$