A street light bulb is fixed on a pole 6m above the level of street. If a woman of height 1.5 m casts a shadow of 3 m, find how far she is away from the base of the pole ?

Let AB be the pole and DE be the woman as shown in the figure below:

Height of pole (AB) = 6 m
and height of a woman (DE) = 1.5 m

Here shadow EF = 3 m

Let BE(Distance of woman from pole) = x meters.

Considering △ABF and △EFD

∠ABF = ∠DEF (Both are equal to 90°)
∠F = ∠F [Common angles]

So, by AA rule of similarity △ABF ~ △EFD. Hence, the ratio of corresponding sides will be equal.

$\therefore \dfrac{BF}{EF} = \dfrac{AB}{DE} \\[1em] \Rightarrow \dfrac{3 + x}{3} = \dfrac{6}{1.5} \\[1em] \Rightarrow \dfrac{3 + x}{3} = \dfrac{60}{15} \\[1em] \Rightarrow 15(3 + x) = 180 \\[1em] \Rightarrow 45 + 15x = 180 \\[1em] \Rightarrow 15x = 180 - 45 \\[1em] \Rightarrow 15x = 135 \\[1em] \Rightarrow x = \dfrac{135}{15} \\[1em] \Rightarrow x = 9.$

Hence, woman is 9 m away from the pole.