\begin{align*} \tan 30^\circ & = {BC \over x} \\ {1 \over \sqrt{3}} & = {BC \over x} \\ x & = \sqrt{3} BC \\ \\ BC & = {1 \over \sqrt{3}} x \times {\sqrt{3} \over \sqrt{3}} \\ & = {\sqrt{3} \over 3} x \\ \\ DC & = {\sqrt{3} \over 3}x \times {1 \over 2} \\ & = {\sqrt{3} \over 6}x \\ \\ \text{By Py} & \text{thagoras theorem,} \\ AD^2 & = DC^2 + CA^2 \\ & = \left( {\sqrt{3} \over 6} x\right)^2 + x^2 \\ & = {1 \over 12}x^2 + x^2 \\ & = {13 \over 12}x^2 \\ AD & = \sqrt{ {13 \over 12}x^2 } \\ & = {\sqrt{13} \over \sqrt{12}} x \\ & = {\sqrt{13} \over 2 \sqrt{3}} x \phantom{000000} [\sqrt{12} = \sqrt{4} \times \sqrt{3} ] \\ & = {\sqrt{13} \over 2\sqrt{3}} x \times {\sqrt{3} \over \sqrt{3}} \\ & = {\sqrt{39} \over 6} x \text{ cm} \end{align*}

\begin{align*} \angle CBA & = 180^\circ - 90^\circ - 30^\circ \\ & = 60^\circ \\ \\ {\sin y^\circ \over BD} & = { \sin 60^\circ \over AD} \phantom{000000} [\text{Sine rule}] \\ AD \sin y^\circ & = BD \sin 60^\circ \\ {\sqrt{39} \over 6} x \sin y^\circ & = \left({\sqrt{3} \over 6} x\right) \left(\sqrt{3} \over 2\right) \\ {\sqrt{39} \over 6} x \sin y^\circ & = {3 \over 12} x \\ {\sqrt{39} \over 6} x \sin y^\circ & = {1 \over 4} x \\ \sin y^\circ & = { {1 \over 4} x \over {\sqrt{39} \over 6} x } \\ & = { {1 \over 4} \over {\sqrt{39} \over 6} } \\ & = {1 \over 4} \div {\sqrt{39} \over 6} \\ & = {1 \over 4} \times {6 \over \sqrt{39}} \\ & = {6 \over 4 \sqrt{39}} \times { \sqrt{39} \over \sqrt{39} } \\ & = {6 \sqrt{39} \over 4(39)} \\ & = {6 \sqrt{39} \over 156} \\ & = {\sqrt{39} \over 26} \\ \\ \therefore y^\circ & = \sin^{-1} \left(\sqrt{39} \over 26\right) \phantom{0} \text{ (Shown)} \end{align*}