Question

In ∆ABC, prove the following:
$\frac{c-b\cos A}{b-c\cos A}=\frac{\cos B}{\cos C}$

Answer 1

Let ABC be any triangle.

$$\begin{gathered} \mathrm{LHS}=\frac{c-b\cos A}{b-c\cos A} \\ =\frac{a\cos B+b\cos A-b\cos A}{a\cos C+c\cos A-c\cos A}\left[\mathrm{Using}\mathrm{projection}\mathrm{formula}: \\ c=a\cos B+b\cos A,b=a\cos C+c\cos A\right] \\ =\frac{a\cos B}{a\cos C} \\ =\frac{\cos B}{\cos C}=\mathrm{RHS} \end{gathered}$$

Hence proved.