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.