Question

in any triangle $\Delta ABC$, with usual notations , prove that $b^2=c^2+a^2-2ca\cos B$

Answer 1

Draw Perpendicular from the vertex A on the the side BC and mark the intersection D .

Let $BD=x$ then $DC=a-x$

Using pythagoras theorem , In $△ABD$ and $△ADC$

$A{D}^2=A{B}^2-B{D}^2$ and $A{D}^2=A{C}^2-D{C}^2$

So $A{B}^2-B{D}^2=A{C}^2-D{C}^2$

$c^2-x^2=$ $b^2-$ ($a-x$)${}^2$

Solving we get $x=\frac{a^2+b^2-c^2}{2a}$ -- $1$

Also In $△ABD$ , $\frac{x}{c}=cosB$ $\to$ $x=c\times CosB$ --$2$

Using equation 1 and 2 , $b^2=c^2+a^2-2ca$ $cosB$