$$\begin{gathered} \mathrm{If} \mathrm{in} \mathrm{a} \mathrm{triangle} \mathrm{ABC}, \mathrm{BC}=5, \mathrm{AC}=4, \mathrm{AB}=3. \\ \mathrm{D}, \mathrm{E} \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{BC} \mathrm{such} \mathrm{that} \\ \mathrm{BD}=\mathrm{DE}=\mathrm{EC}, \angle \mathrm{CAE}=\mathrm{\theta }. \mathrm{Then} \end{gathered}$$

$$\begin{gathered} \mathrm{BC}=5, \mathrm{AC}=4, \mathrm{AB}=3; \mathrm{Triangle} \mathrm{ABC} \mathrm{is} \mathrm{right} \mathrm{angle} \mathrm{with} \angle \mathrm{A}={90}^0 \\ \mathrm{Let} \mathrm{A}=\left(0, 0\right), \mathrm{C}=\left(0, 4\right), \mathrm{B}=\left(3, 0\right) \\ \therefore \mathrm{D}=\left(2, \frac{4}{3}\right), \mathrm{E}=\left(1, 8/3\right) \mathrm{AE}=\sqrt{1+\frac{64}{9}}=\sqrt{\frac{73}{9}}\Rightarrow A{E}^2=\frac{73}{9} \\ and \mathrm{AD}=\sqrt{4+\frac{16}{9}}=\sqrt{\frac{52}{9}}=\frac{2\sqrt{13}}{3} \\ In∆AEB \frac{10/3}{\sin \left(90-\theta \right)}=\frac{AE}{\sin B}, \\ In ∆AEC,\frac{5/3}{\sin \theta }=\frac{AE}{\sin C}\Rightarrow \frac{10\sin B}{3\cos \theta }=\frac{5\sin C}{3\sin \theta } \\ \Rightarrow \tan \theta =\frac{\sin C}{2\sin B}=\frac{3/5}{2(4/5)}\Rightarrow \tan \mathrm{\theta }=3/8 \end{gathered}$$