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

Solution :

Given : Let AC be a tower casts a shadow BC = 28 m and DF = 6 m be a pole casts a shadow EF = 4mtriangles

To Find : Height of the tower

Procedure : Now, In triangle ABC and DEF

\(\angle\) ACB = \(\angle\) DEF          (each 90)

At the same time the rays of sun have same inclination

\(\implies\)   \(\angle\) ABC = \(\angle\) DEF

By AA similarity,

\(\triangle\) ABC ~ \(\triangle\) DEF

\(\implies\)  \(AC\over DF\) = \(BC\over EF\)

\(\implies\)  \(AC\over 6\) = \(28\over 4\)

\(\implies\)  AC = 42 m

Hence, the height of tower is 42 m.

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