= Solution
Use the inverse standard <Lorentz boost>, $t=\gamma_v(t'+vx'/c^2)$ and $x=\gamma_v(x'+vt')$. Label two rod endpoints by a separation $\Delta x'=L$; their world lines satisfy $y'=ut'$. A rod's orientation in $S$ must compare endpoint events simultaneous in $S$, not in $S'$. Consequently
$$
0=\Delta t=\gamma_v\left(\Delta t'+\frac{vL}{c^2}\right),\qquad
\Delta t'=-\frac{vL}{c^2}.
$$
Then
$$
\Delta x=\gamma_v(L+v\Delta t')=\frac L{\gamma_v},\qquad
\Delta y=u\Delta t'=-\frac{uvL}{c^2}.
$$
The signed counterclockwise angle therefore satisfies
$$
\boxed{\tan\theta=-\gamma_v\frac{uv}{c^2}.}
$$
For positive $u,v$, the rod slopes downward to the right, so the acute clockwise angle is $\arctan(\gamma_vuv/c^2)$. This is a <relativity of simultaneity> effect in addition to the longitudinal <length contraction>; the rod is not being mechanically bent by the coordinate change.
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