Question

In ∆ABC, prove the following:

$\frac{c-b\cos A}{b-c\cos A}=\frac{\cos B}{\cos C}$

Answer 1

$$\begin{gathered} \mathrm{Consider} \\ \frac{c-b\cos A}{b-c\cos A} \end{gathered}$$
$\mathrm{On}\mathrm{using}\mathrm{the}\mathrm{projection}\mathrm{formula},c=a\cos B+b\cos A,b=c\cos A+a\cos C,\mathrm{we}\mathrm{get}:$

$$\begin{gathered} \frac{c-b\cos A}{b-c\cos A}=\frac{a\cos B+b\cos A-b\cos A}{c\cos A+a\cos C-c\cos A} \\ =\frac{a\cos B}{a\cos C} \\ =\frac{\cos B}{\cos C} \end{gathered}$$

Hence proved.