Banked curve

Tilt the surface and the normal force tilts with it, since it always points perpendicular to the surface. Its horizontal component then aims at the center of the turn, which is where the net force has to point.

AP Physics 1 EK 2.9.A.2.ii, repeated as C: Mechanics EK 2.10.A.2.ii, says components of the static friction force and the normal force can contribute to the net force producing centripetal acceleration of an object traveling in a circle on a banked surface.

Read the AP Physics 1 boundary statement whole. It says AP Physics 1 only expects students to quantitatively analyze banked curves in which no friction is required to maintain uniform circular motion, and that analysis of situations in which friction is required on a banked curve is limited to qualitative descriptions. So the numbers you are asked for are the frictionless case; the friction case still needs a written answer. The C: Mechanics CED prints no equivalent restriction.

The frictionless geometry, derived rather than printed. Nothing accelerates vertically, so $N\cos\theta = mg$. Horizontally the net force is centripetal, so $N\sin\theta = mv^2/r$. Divide and the mass cancels: $\tan\theta = v^2/(rg)$. One bank angle suits one speed, whatever the vehicle weighs.

The normal force is not $mg\cos\theta$ here. That result belongs to an object resting on an incline. On a banked curve the vertical acceleration is zero while the object accelerates horizontally, giving $N = mg/\cos\theta$, larger than the weight rather than smaller. No arrow labelled centripetal force belongs on the diagram either; the centripetal force guide covers that.