G3.2 In figure $1, C$ lies on $A D, A B=B D=1 \mathrm{~cm}, \angle A B C=90^{\circ}$ and $\angle C B D=30^{\circ}$. If $C D=b \mathrm{~cm}$, find the value of $b$.
$A B=B D=1 \mathrm{~cm}, \triangle A B D$ is isosceles.
$\angle B A D=\angle B D A=\left(180^{\circ}-90^{\circ}-30^{\circ}\right) \div 2=30^{\circ}(\angle \mathrm{s}$ sum of isosceles $\triangle)$ $\triangle B C D$ is also isosceles.
$C D=b \mathrm{~cm}=B C=A B \tan \angle B A D=1 \tan 30^{\circ} \mathrm{cm}=\frac{1}{\sqrt{3}} \mathrm{~cm}$