A 1.5m tall boy is standing …

Let AE is the Length of the building. 

So AE = 30

Again BE = DF = 1.5

AB = AE - BE = 30 - 1.5 = 28.5

Now in triangle ABC,

tan60 = {tex}\frac{AB}{BC}{/tex}

{tex}\Rightarrow {/tex} √3 = {tex}\frac{28.5}{BC}{/tex}

{tex}\Rightarrow {/tex} BC = {tex}\frac{28.5}{\sqrt 3}{/tex}

Again in triangle ABD

tan30 = {tex}\frac{AB}{BD}{/tex}
{tex}\sqrt{1}{\sqrt3}=\frac{28.5}{BD}{/tex}

{tex}\Rightarrow {/tex} BD = 28.5{tex}\times{/tex}√3

{tex}\Rightarrow {/tex} BC + CD = 28.5√3

{tex}\Rightarrow {/tex}  28.5/√3 + CD = 28.5√3

{tex}\Rightarrow {/tex} CD = {tex}\frac{28.5}{\sqrt 3}{/tex}- {tex}\frac{28.5}{\sqrt 3}{/tex}

{tex}\Rightarrow {/tex} CD  = {tex}\frac{28.5\times3-28.5}{\sqrt3}{/tex}

{tex}\Rightarrow {/tex} CD = {tex}\frac{28.5(3-1)}{\sqrt3}{/tex}

{tex}\Rightarrow {/tex} CD = {tex}\frac{(28.5\times 2)}{\sqrt3}{/tex}

{tex}\Rightarrow {/tex} CD = {tex}\frac{(57)}{\sqrt3}{/tex}

{tex}\Rightarrow {/tex} CD = {tex}\frac{(57\sqrt3)}{\sqrt3\times\sqrt3}{/tex} (Multiply √3 in numerator and denominator)

{tex}\Rightarrow {/tex} CD = {tex}\frac{57\sqrt3}{3}{/tex}

{tex}\Rightarrow {/tex} CD = 19√3

The distance he walked towards the building is 19√3  m