Giải thích các bước giải:

a.Ta có: $\hat E=30^o$

$\sin E=\dfrac{BD}{DE}$

$\to DE=\dfrac{BD}{\sin E}=\dfrac{x}{\sin30^o}=2x$

b.Ta có: $BD=1.5$

$\to DE=3$

$\to BE=\sqrt{DE^2-DB^2}=\dfrac{3\sqrt3}2$

Ta có; $BA=2BD=3$

$\tan C=\dfrac{AB}{BC}\to BC=\dfrac{AB}{\tan C}=\dfrac{3}{\tan60^o}=\sqrt3$

$\to CE=BE+BC=\dfrac{3\sqrt3}2+\sqrt3=\dfrac{5\sqrt3}2$