Two posts are k meter apart …

Let AB and CD be the two posts such that AB = 2CD.
and the distance between the two posts given as k
Let M be the mid-point of CA. Let {tex}\angle{/tex}CMD = {tex}\theta{/tex}
and {tex}\angle{/tex}AMB = 90° -{tex}\theta{/tex}
Clearly, {tex}\mathrm { CM } = M A = \frac { 1 } { 2 } k{/tex}
Let CD = h, then AB = 2h
Now, {tex}\frac { A B } { A M } = \tan \left( 90 ^ { \circ } - \theta \right) = \cot \theta{/tex} 
{tex}\Rightarrow \quad \frac { 2 h } { \left( \frac { k } { 2 } \right) } = \cot \theta{/tex} 
{tex}\Rightarrow \quad \cot \theta = \frac { 4 h } { k }{/tex}  ...(i)
Also in {tex}\triangle {/tex}CMD, {tex}\frac { C D } { C M } = \tan \theta{/tex} 
{tex}\Rightarrow \quad \frac { h } { \frac { k } { 2 } } = \tan \theta{/tex}
{tex}\Rightarrow \quad \tan \theta = \frac { 2 h } { k }{/tex}  ........(ii)
Multiplying (i) and (ii), {tex}\frac { 4 h } { k } \times \frac { 2 h } { k } = 1{/tex}
{tex}\therefore \quad h ^ { 2 } = \frac { k ^ { 2 } } { 8 }{/tex} 
{tex}\Rightarrow \quad h = \frac { k } { 2 \sqrt { 2 } } = \frac { k \sqrt { 2 } } { 4 }{/tex}