Question

In a $△ABC,\frac{b+c}{a}=\frac{\cos B+\cos C}{1-\cos A}$

• A. True
• B. False

Answer 1

The correct option is A True

From projection rule, we have
$b=c\cos A+a\cos C,c=a\cos B+b\cos A$
$\Rightarrow \frac{b+c}{a}=\frac{c\cos A+a\cos C+a\cos B+b\cos A}{a}$
$\Rightarrow \frac{b+c}{a}=\frac{(b+c)\cos A+a(\cos B+\cos C)}{a}$
$\Rightarrow \frac{b+c}{a}-\frac{(b+c)\cos A}{a}=\frac{a(\cos B+\cos C)}{a}$
$\Rightarrow (b+c)(1-\cos A)=a(\cos B+\cos C)$
$\Rightarrow b+c=\frac{a(\cos B+\cos C)}{(1-\cos A)}$
Divide both sides by $a$ we get
$\Rightarrow \frac{b+c}{a}=\frac{(\cos B+\cos C)}{(1-\cos A)}$