Question
In any $\triangle ABC$, with usual notations, prove that $b^2=c^2+a^2-2 c a \cos B$.
✓
Answer
Let us take the angle $B$ of$ \triangle ABC$ in standard position,
$i.e. B$ as origin, $X-$axis along the line $BC$ and the $Y-$axis perpendicular to the line $BC$.
In the two figures,$ \angle B$ is shown as acute in one and obtuse in the other.
$\because l(BC) = a$
$\therefore C ≡ (a, 0$)
Let $A ≡ (x, y)$
Since $l (BA) = c,$ we have
$\cos B =\frac{x}{c}$ and $\sin B =\frac{y}{c}$
$\therefore x = c \cos B$ and $y = c \sin B$
$\therefore A ≡ (c \cos B, c \sin B)$
$\therefore$ By the distance formula
$b^2 = AC^2 = (a – c \cos B)^2 + (0 – c \sin B)^2$
$= a^2 – 2ca \cos B + c^2 \cos^2B + c^2 \sin^2B$
$= c^2(\cos^2B + \sin^2B) + a^2 – 2ca \cos B$
$\therefore b^2 = c^2 + a^2 – 2ca \cos B.$
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