Question

A vertical pole of length 6 m casts a shadow 4 m long on the ground. At the same time, a tower casts a shadow 28 m long .
Find the height of the tower.

Answer 1

Given :
AB = 6 m , BC = 4 m and YZ = 28 m

To find :XY


$$\begin{gathered} \text{In}△\mathrm{ABC}\text{and}△\mathrm{XYZ}, \\ \angle \mathrm{ABC}\cong \angle \mathrm{XYZ}(90°\mathrm{each}) \\ \text{The}\mathrm{angle}\mathrm{of}\mathrm{elevation}\mathrm{for}\mathrm{the}\mathrm{person}\mathrm{standing}\mathrm{at}\mathrm{C}\text{and}\text{for}\mathrm{the}\mathrm{person}\mathrm{standing} \\ \mathrm{at}\mathrm{Z}\mathrm{would}\mathrm{be}\mathrm{the}\mathrm{same}. \\ \therefore \angle \mathrm{ACB}\cong \angle \mathrm{XZY} \\ \mathrm{Therefore},\mathrm{by}\text{the}\mathrm{AA}\mathrm{similarity}\mathrm{test}, \\ △\mathrm{ABC}~△\mathrm{XYZ} \\ \mathrm{By}\mathrm{the}\mathrm{property}\mathrm{of}\mathrm{similarity}, \\ \frac{\mathrm{AB}}{\mathrm{XY}}=\frac{\mathrm{BC}}{\mathrm{YZ}} \\ \Rightarrow \frac{6}{\mathrm{XY}}=\frac{4}{28}\Rightarrow \mathrm{XY}=\frac{28\times 6}{4}=42m \end{gathered}$$
Hence , the height of the tower is 42 m.