${x \over d} = \tan {30^o} \Rightarrow {x \over d} = {1 \over {\sqrt 3 }} \Rightarrow d = \sqrt 3 x$ and ${{x + 30} \over d} = \tan {45^o}$ $\Rightarrow$ d = x + 30 $d = {d \over {\sqrt 3 }} + 30 \Rightarrow \left( {1 - {1 \over {\sqrt 3 }}} \right)d = 30$ $\Rightarrow d = {{30\sqrt 3 } \over {\sqrt 3 - 1}} \Rightarrow d = {{30\sqrt 3 (\sqrt 3 + 1)} \over 2} = 15\sqrt 3 (\sqrt 3 + 1)$ $\Rightarrow$ d = 15 ( 3 + $\sqrt 3$)