In triangle $ABC,$
\[(b \sin C)(b \cos C + c \cos B) = 42.\]Compute the area of triangle $ABC.$
By the Law of Cosines,
\begin{align*}
b \cos C + c \cos B &= b \cdot \frac{a^2 + b^2 - c^2}{2ab} + c \cdot \frac{a^2 + c^2 - b^2}{2ac} \\
&= \frac{a^2 + b^2 - c^2}{2a} + \frac{a^2 + c^2 - b^2}{2a} \\
&= \frac{2a^2}{2a} = a,
\end{align*}so $ab \sin C = 42.$

Then the area of triangle $ABC$ is
\[\frac{1}{2} ab \sin C = \boxed{21}.\]