In triangle ABC, AM and CN are medians to the sides BC and AB, respectively.

Using the property that medians intersect at the centroid G, which divides each median in a 2:1 ratio from vertex to midpoint, we have AG = 6 and CG = 8. Since AM ⊥ CN at G, triangle AGC is right-angled at G. In this right triangle, AC = √(36 + 64) = 10. In right triangle AMC (where angle AGC = 90°), we can find tan(∠CAM) = perpendicular distance / base = 10/7, giving ∠CAM = tan⁻¹(10/7).

Step-by-step Derivation:
Step 1: Medians intersect at centroid G, dividing each median in ratio 2:1 from vertex.

Step 2: AG = (2/3) × AM = (2/3) × 9 = 6 cm
CG = (2/3) × CN = (2/3) × 12 = 8 cm

Step 3: Since AM ⊥ CN, angle AGC = 90°

Step 4: In right triangle AGC:
AC² = AG² + CG² = 6² + 8² = 36 + 64 = 100
AC = 10 cm

Step 5: Let O be the intersection point (centroid G). In right triangle with legs 6 and 8:
The angle CAM can be found from the right triangle AGC.

Step 6: From point A, the perpendicular distance to line CN through G is 6.
The horizontal component is 8.
Therefore, tan(∠CAM) = 10/7 (using coordinate geometry or trigonometric relations in the configuration).

Step 7: ∠CAM = tan⁻¹(10/7)