From the top of an $\displaystyle 8$ m high building, the angle of elevation of the top of a cable tower is $\displaystyle 60^{\circ}$ and the angle of depression of its foot is $\displaystyle 45^{\circ}$. Determine the height of the tower. (Take $\displaystyle \sqrt{3}=1.732$ ).

Let $\displaystyle A B=$ height of building $\displaystyle =8 \mathrm{~m}$ Let $\displaystyle C D=$ height of tower $\displaystyle =h \mathrm{~m}$ \[\begin{array}{l} \angle D B E=60^{\circ} \\ \angle A C B=\angle E B C=45^{\circ} \\ A C=B E=y(\mathrm{let}) \end{array} \]In right $\displaystyle \triangle A B C$, \[\begin{array}{l} \tan 45^{\circ}=\frac{8}{A C} \\ 1=\frac{8}{A C} \\ \Rightarrow A C=8 \mathrm{~m} \Rightarrow y=8 \mathrm{~m} \end{array} \]In right $\displaystyle \triangle B D E, \tan 60^{\circ}=\frac{h-8}{B E}$ \[\begin{array}{l} \sqrt{3}=\frac{h-8}{y} \Rightarrow \sqrt{3} y=h-8 \\ \sqrt{3}(8)=h-8 \\ h=8 \sqrt{3}+8=8(\sqrt{3}+1) \\ h=8(1 \cdot 732+1)=8(2 \cdot 732)=21 \cdot 856 \mathrm{~m} \end{array} \] ∴ Height of tower $\displaystyle =21.856 \mathrm{~m}$