如图, 在 $\triangle A B C$ 中, $\cos B=\frac{\sqrt{2}}{2}, \sin C=\frac{3}{5}, A C=5$, 则 $\triangle A B C$ 的面积是 ( )

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过点 $\mathrm{A}$ 作 $\mathrm{AD} \perp \mathrm{BC}$,

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$\because \triangle \mathrm{ABC}$ 中, $\cos \mathrm{B}=\frac{\sqrt{2}}{2}, \sin \mathrm{C}=\frac{3}{5}, \mathrm{AC}=5$,

$\therefore \cos \mathrm{B}=\frac{\sqrt{2}}{2}=\frac{B D}{A B}, \therefore \angle \mathrm{B}=45^{\circ}, \because \sin \mathrm{C}=\frac{3}{5}=\frac{A D}{A C}=\frac{A D}{5}$,

$\therefore \mathrm{AD}=3, \quad \therefore \mathrm{CD}=\sqrt{5^{2}-3^{2}}=4, \quad \therefore \mathrm{BD}=3$,

则 $\triangle \mathrm{ABC}$ 的面积是: $\frac{1}{2} \times \mathrm{AD} \times \mathrm{BC}=\frac{1}{2} \times 3 \times(3+4)=\frac{21}{2}$. 故选: $\mathrm{A}$.