Question

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

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Solution

## Length of the vertical pole $=6m$Shadow of the pole $=4m$Length of the shadow of the tower $=28m$Let the height of the tower$=hm$From $\Delta ABCand\Delta DEF,$$\angle C=\angle E$ (angular elevation of the sum)$\angle B=\angle F=90°$$\therefore \Delta ABC~\Delta DEF$ (AA similarity criterion)Hence, $\frac{AB}{DF}=\frac{BC}{EF}$ (If two triangles are similar corresponding sides are proportional)So we can represent it in the equation as$\frac{6}{h}=\frac{4}{28}\phantom{\rule{0ex}{0ex}}⇒h=\frac{\left(6×28\right)}{4}\phantom{\rule{0ex}{0ex}}⇒h=6×7\phantom{\rule{0ex}{0ex}}⇒h=42m$Hence, the height of the tower is $42m.$

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