## Precalculus (6th Edition)

$10\sqrt {2}$
From general properties of triangle $\angle B=180-\left( \angle A+\angle C\right) =180-\left( 105+45\right) =30$ From the law of sines $\dfrac {a}{\sin 45}=\dfrac {10}{\sin B}\Rightarrow a=\dfrac {10\sin 45}{\sin 30}=\dfrac {10\times \dfrac {\sqrt {2}}{2}}{\dfrac {1}{2}}=10\sqrt {2}$