Sum of Sides of Triangle in terms of Circumradius and Half Angle Cosines
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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:
- $a + b + c = 8 R \cos \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2$
where $R$ denotes the circumradius of $\triangle ABC$.
Proof
From Law of Sines:
- $\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C} = 2 R$
Hence:
\(\ds a\) | \(=\) | \(\ds 2 R \sin A\) | ||||||||||||
\(\ds b\) | \(=\) | \(\ds 2 R \sin B\) | ||||||||||||
\(\ds c\) | \(=\) | \(\ds 2 R \sin C\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a + b + c\) | \(=\) | \(\ds 2 R \paren {\sin A + \sin B + \sin C}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 R \paren {4 \cos \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2}\) | Sum of Sines of Angles in Triangle | |||||||||||
\(\ds \) | \(=\) | \(\ds 8 R \cos \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2\) |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Exercises $\text {XXXII}$: $14$.