Length of Side of Triangle as Cosines of Angles

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Theorem

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Then:

$c = a \cos B + b \cos A$


Proof

Side-of-Triangle-by-Cosines.png

Let a perpendicular be dropped from $C$ to $AB$ to meet $AB$ at $D$.

There are two cases:


Foot of Perpendicular is on $AB$

See the diagram on the left.

\(\ds c\) \(=\) \(\ds AB\)
\(\ds \) \(=\) \(\ds DB + AD\)
\(\ds \) \(=\) \(\ds BC \cos B + AC \cos A\) Definition of Cosine of Angle
\(\ds \) \(=\) \(\ds a \cos B + b \cos A\)

$\Box$


Foot of Perpendicular is outside $AB$

See the diagram on the right.

\(\ds c\) \(=\) \(\ds AB\)
\(\ds \) \(=\) \(\ds AD - BD\)
\(\ds \) \(=\) \(\ds AC \cos A - BC \cos \angle CBD\) Definition of Cosine of Angle
\(\ds \) \(=\) \(\ds AC \cos A - BC \map \cos {180 \degrees - \angle ABC}\)
\(\ds \) \(=\) \(\ds AC \cos A + BC \cos \angle ABC\) Cosine of Supplementary Angle
\(\ds \) \(=\) \(\ds a \cos B + b \cos A\)

$\blacksquare$


Sources

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