Official solution
\[\tan {15^o} = {5 \over d} \Rightarrow d = {5 \over {\tan {{15}^o}}}\]
\[ \Rightarrow {{5\left( {\sqrt 3 + 1} \right)} \over {\sqrt 3 - 1}} \Rightarrow {{5\left( {4 + 2\sqrt 3 } \right)} \over 2}\]
\[ = 5\left( {2 + \sqrt 3 } \right)\]