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kuing
posted 2023-11-13 15:58
记圆半径为 `R`,则
\begin{align*}
a&=BC=\sqrt{n^2+r^2-2nr\cos\theta},\\
b&=AC=\sqrt{m^2+r^2+2mr\cos\theta},\\
\S{ABC}&=\frac12ch_c=\frac12cr\sin\theta,\\
R&=\frac{abc}{4\S{ABC}}=\frac{\sqrt{n^2+r^2-2nr\cos\theta}\sqrt{m^2+r^2+2mr\cos\theta}}{2r\sin\theta}.
\end{align*} |
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