Exradius of Triangle in Terms of Circumradius/Proof 2
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Theorem
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $\rho_a$ be the exradius of $\triangle ABC$ with respect to $a$.
Let $R$ be the circumradius of $\triangle ABC$.
Then:
- $\rho_a = 4 R \sin \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2$
Proof
Let $r$ denote the inradius of $\triangle ABC$.
We have:
\(\ds r\) | \(=\) | \(\ds 4 R \sin \dfrac A 2 \sin \dfrac B 2 \sin \dfrac C 2\) | Inradius in Terms of Circumradius | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \rho_a\) | \(=\) | \(\ds 4 R \sin \dfrac A 2 \map \sin {\dfrac {180 \degrees - B} 2} \map \sin {\dfrac {180 \degrees - C} 2}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 R \sin \dfrac A 2 \map \sin {90 \degrees - \dfrac B 2} \map \sin {90 \degrees - \dfrac C 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4 R \sin \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2\) | Sine of Complement equals Cosine |
This article, or a section of it, needs explaining. In particular: Why the second line? From the book it says "It should be noted that this formula can be deduced from Inradius in Terms of Circumradius by writing $\paren {180 \degrees - B}$ for $B$ and $\paren {180 \degrees - C}$ for $C$" but plenty of work is needed to show why this is valid. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The ex-circles