In ${\mathrm{BCl}}_3$ molecule, the $\mathrm{Cl}-\mathrm{B}-\mathrm{Cl}$ bond angle is

Boron is the central atom. It has three valence electrons. These electrons combine with the electrons of chlorine atoms and form three B-Cl bonds. So, there is ${\mathrm{sp}}^2$hybridization and trigonal planar shape. The bond angle in the molecule is ${120}^{\mathrm{o}}$. 
${\mathrm{BCl}}_3: \mathrm{B}\to {\mathrm{SP}}^2\to \mathrm{Trigonal} \mathrm{planer}$