Official solution
Let a \[\Delta\]ABC having C = 90\[^\circ\] and A = \[\theta\]
\[{{\sin \theta } \over a} = {{\cos \theta } \over b} = {1 \over c}\] ..... (i)
Also for triangle of reciprocals
\[\cos A = {{{{\left( {{1 \over c}} \right)}^2} + {{\left( {{1 \over b}} \right)}^2} - {{\left( {{1 \over a}} \right)}^2}} \over {2\left( {{1 \over c}} \right)\left( {{1 \over b}} \right)}}\]
\[{1 \over {{c^2}}} + {1 \over {{{(c\cos \theta )}^2}}} = {1 \over {{{(c\sin \theta )}^2}}}\]
\[ \Rightarrow 1 + {\sec ^2}\theta = \cos e{c^2}\theta \]
\[ \Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {4{{\sin }^2}\theta {{\cos }^2}\theta }}\]
\[ \Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {{{\sin }^2}2\theta }}\]
\[ \Rightarrow 1 - {\cos ^2}2\theta = 4\cos 2\theta \]
\[{\cos ^2}2\theta + 4\cos 2\theta - 1 = 0\]
\[\cos 2\theta = {{ - 4 \pm \sqrt {16 + 4} } \over 2}\]
\[\cos 2\theta = - 2 \pm \sqrt 5 \]
\[\cos 2\theta = \sqrt 5 - 2 = 1 - 2{\sin ^2}\theta \]
\[ \Rightarrow 2{\sin ^2}\theta = 3 - \sqrt 5 \]
\[ \Rightarrow {\sin ^2}\theta = {{3 - \sqrt 5 } \over 2}\]
\[ \Rightarrow \sin \theta = {{\sqrt 5 - 1} \over 2}\]