A girl of height 90cm is …

We have,

Height of girl = 90 cm = 0.9 m
Height of lamp-post = 3.6 m
Speed of girl = 1.2 m/sec
Time taken = 4 sec.
{tex}\therefore{/tex} Distance moved by girl (CQ) = Speed {tex}\times{/tex} Time
= 1.2 {tex}\times{/tex} 4
= 4.8 m
Let length of shadow (AC) = x cm
In {tex}\Delta{/tex}ABC and {tex}\Delta{/tex}APQ
{tex}\angle ACB = \angle AQP{/tex} [Each 90°]
{tex}\angle BAC = \angle PAQ{/tex} [Common]
Then, {tex}\Delta{/tex}ABC ~ {tex}\Delta{/tex}APQ [By AA similarity]
{tex}\therefore \frac{{AC}}{{AQ}} = \frac{{BC}}{{PQ}}{/tex} [Corresponding parts of similar {tex}\Delta{/tex} are proportional]
{tex} \Rightarrow \frac{x}{{x + 4.8}} = \frac{{0.9}}{{3.6}}{/tex}
{tex}\Rightarrow \frac{x}{{x + 4.8}} = \frac{1}{4}{/tex}
{tex}\Rightarrow{/tex} 4x = x + 4.8
{tex}\Rightarrow{/tex} 4x - x = 4.8
{tex}\Rightarrow{/tex} 3x = 4.8
{tex}\Rightarrow x = \frac{{4.8}}{3} = 1.6m{/tex}
{tex}\therefore{/tex} Length of shadow = 1.6 m