Let AB = Lamp post
CD = girl
DE = shadow of the girl,
BD = 1.2 × 4 = 4.8 m
Let shadow of girl,
DE = ‘a’ m
In ∆ABE and ∆CDE
∠B = ∆(each 90°)
∠E = ∠E (comman angle)
∆ABE ~ ∆CDE by (AA criterion)
therefore, BE/DE = AB/CD ⇒ (4.8 + a)/a = 3.6/0.9
⇒ 4.8 + a = 4a
⇒ 3 a = 4.8
⇒ a = 1.6 m
Thus after 4 sec. length of girl’s shadow will be 1.6 m