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[几何] ACD的面积范围

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Zach Posted at 2024-7-30 08:52:48 |Read mode
ACD的面积范围?
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战巡 Posted at 2024-7-30 12:33:49
这不是个范围,只是两个特定的解而已

令$AB=h$,则
\[AC=\sqrt{h^2+3^2}, AD=\sqrt{h^2+20^2}\]

\[AC^2+AD^2-17^2=2AC\cdot AD\cos(45\du)\]
解得
\[h=5,h=12\]
于是(1)、(5)

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