第二題

【劉丞諭提供】
自\(P\)作\(\overline {PD}\) \(\perp\) \(\overline {AB}\)於\(D\)，自\(P\)作\(\overline {PE}\) \(\perp\) \(\overline {AQ}\)於\(E\)，如圖﹒
設\( \overline{BP}= \overline{PQ} = \overline{CQ}=\sqrt{2} \),\(\overline{AE}=x\).
易知\( \overline{BP}= \overline{PD} = {1} \),\(\overline{AD}={2}\),

所以\(\overline{AP}=\sqrt{5}\)，同理\(\overline{AQ}=\sqrt{5}\).
在\(\Delta {ABC}\)中，\({\overline{PE}}^{2}=(\sqrt{2})^{2}-({\sqrt{5}-x})^{2} =(\sqrt{5})^{2}-{x}^{2}\).
解得\(x=\frac{4}{\sqrt{5}} \)即\(\overline{AE}=\frac{4}{\sqrt{5}} \)，\(\overline{PE}=\frac{3}{\sqrt{5}} \)，所以\({\rm tan} \angle{PAQ}=\frac{\frac{3}{\sqrt(5)}}{\frac{4}{\sqrt({5}}}=\frac{3}{4}\).