Answer:
Let AB be tower of height h From \[\Delta \,ABC,\frac{AB}{BC}=\tan \,\,45{}^\circ \] \[h=BC\] From \[\Delta \,\,ABD,\frac{AB}{BD}=\tan \,\,30{}^\circ \] \[h=\frac{BD}{\sqrt{3}}\] or \[BD=\sqrt{3}\,h\] \[CD=BD-BC\] \[\Rightarrow \] \[=\sqrt{3}\,h-h=\left( \sqrt{3}-1 \right)h\] \[\Rightarrow \] \[20\,m=\left( \sqrt{3}-1 \right)h\] \[\Rightarrow \] \[h=\frac{20}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{20\left( \sqrt{3}+1 \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{(1)}^{2}}}\] \[=\frac{20\left( \sqrt{3}+1 \right)}{2}=10\left( \sqrt{3}+1 \right)m\]
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