# 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 = 4m

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.