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,