According to ques,

: In ∆ABC, 

\[\angle C\text{ }=\text{ }{{105}^{o}},\]

\[~\angle B\text{ }=\text{ }{{45}^{o}},\]

\[~a\text{ }=\text{ }2\]

Since,

\[\angle A\text{ }+~\angle B\text{ }+~\angle C\text{ }=\text{ }180{}^\circ \]

\[\angle A\text{ }=\text{ }180{}^\circ \text{ }~\angle B\text{ }~\angle C\]

From above mentioned values, we have,

\[\angle A\text{ }=\text{ }180{}^\circ \text{ }\text{ }45{}^\circ \text{ }\text{ }105{}^\circ \]

\[\angle A\text{ }=\text{ }30{}^\circ \]

By sine rule, we have,