Solution:

Using the formula $\tan \theta = \frac{v^2}{r g}$ and given values $v = 10$ m/s, $r = 10$ m, and $g = 10$ m/s², we get:

$$\tan \theta = \frac{10^2}{10 \times 10} = 1$$

Thus, $\theta = \tan^{-1}(1) = 45^\circ$. The cyclist leans at 45° or π/4 with the vertical.