in the given figure ABCD is a rectangle. with AD = 8cm and CD =12 cm. Line segment CE is drawn making an angle 60 with AB, intersecting AB in E. Find the length of CE and BE.
 

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$$\begin{gathered} \mathrm{As} \mathrm{opposite} \mathrm{sides} \mathrm{of} \mathrm{rectangle} \mathrm{are} \mathrm{equal}, \mathrm{hence} \mathrm{we} \mathrm{have}: \\ \mathrm{AB}=\mathrm{CD}=12 \mathrm{cm} \\ \mathrm{BC}=\mathrm{AD}=8 \mathrm{cm} \\ \angle \mathrm{BEC}=60 \\ \mathrm{In} \mathrm{right} \mathrm{angled} ∆\mathrm{BEC} \\ \tan 60=\frac{\mathrm{BC}}{\mathrm{BE}} \\ \Rightarrow \mathrm{BE}=\frac{\mathrm{BC}}{\tan 60} \\ \Rightarrow \mathrm{BE}=\frac{8}{\sqrt{3}} \\ \mathbf{\Rightarrow }\mathbf{BE}\mathbf{=}\frac{\mathbf{8}\sqrt{\mathbf{3}}}{\mathbf{3}}\mathbf{ }\mathbf{cm} \\ \sin 60=\frac{\mathrm{BC}}{\mathrm{CE}} \\ \Rightarrow \mathrm{CE}=\frac{\mathrm{BC}}{\sin 60} \\ \Rightarrow \mathrm{CE}=\frac{8}{{\displaystyle \frac{\sqrt{3}}{2}}} \\ \Rightarrow \mathrm{CE}=\frac{16}{{\displaystyle \sqrt{3}}} \\ \mathbf{\Rightarrow }\mathbf{CE}\mathbf{=}\frac{\mathbf{16}\sqrt{\mathbf{3}}}{{\displaystyle 3}}\mathbf{ }\mathbf{cm} \end{gathered}$$

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