Question

Show that in $\Delta ABC,a=b\cos C+c\cos B$

Answer 1

To prove in $△ABC$

Main formula $\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow a=b\cos C+c\cos B$

This is called projection formula :

In the RHS we have :-

$b\cos c+c\cos b$

we know cosine formula

$\Rightarrow \cos A=\frac{b^2+c^2-b^2}{2bc}$

using cosine formula,

$=b\frac{(b^2+a^2-c^2)}{2ab}+c\frac{(c^2+a^2-b^2)}{2ac}$

$=\frac{b^2+a^2-c^2}{2a}+\frac{c^2+a^2-b^2}{2a}$

$=\frac{2a^2}{2a}=a$

on LHS we have a LHS $=$ RHS

Hence, solved.