設 $A$ 為原點，$B = (a, 0)$，$C = (0, a)$。 $P$、$Q$ 為 $\overline{BC}$ 的三等分點： $$P = B + \dfrac{1}{3}(C - B) = \left(\dfrac{2a}{3},\ \dfrac{a}{3}\right)$$ $$Q = B + \dfrac{2}{3}(C - B) = \left(\dfrac{a}{3},\ \dfrac{2a}{3}\right)$$ $\angle BAP$ 的正切值：$\tan \angle BAP = \dfrac{a/3}{2a/3} = \dfrac{1}{2}$ $\angle BAQ$ 的正切值：$\tan \angle BAQ = \dfrac{2a/3}{a/3} = 2$ $\angle PAQ = \angle BAQ - \angle BAP$，利用正切差角公式： $$\tan \angle PAQ = \tan(\angle BAQ - \angle BAP) = \dfrac{2 - \dfrac{1}{2}}{1 + 2 \times \dfrac{1}{2}} = \dfrac{\dfrac{3}{2}}{2} = \dfrac{3}{4}$$