If a road is banked at an angle $\theta$, and the cyclist is leaning inward at an angle $\theta'$ to the vertical, for an ideal turn at the optimum speed, what is the relationship between the angles?

For an ideal turn at the optimum speed on a banked road, no frictional force is required. The necessary centripetal force is provided entirely by the horizontal component of the normal force.

The optimum speed $v$ for a road banked at an angle $\theta$ is given by:
$\tan \theta = \dfrac{v^2}{rg}$

When a cyclist takes a turn of radius $r$ at speed $v$, they must lean inward at an angle $\theta'$ to the vertical such that the net force passes through their center of mass. This angle of lean is given by:
$\tan \theta' = \dfrac{v^2}{rg}$

Equating the two expressions, we get:
$\tan \theta' = \tan \theta$

$\Rightarrow \theta' = \theta$

Thus, the cyclist aligns their body along the normal to the banked road.

Answer: $\theta' = \theta$