1. The problem asks to calculate the value of $n$ and the interior angles of two regular polygons with sides $n-1$ and $n$ respectively, given that the sum of their interior angles is in the ratio 2:3. 2. Recall the formula for the sum of interior angles of a polygon with $k$ sides: $$\text{Sum of interior angles} = (k-2) \times 180^\circ$$ 3. Let the sum of interior angles of the polygon with $n-1$ sides be $S_1$ and that of the polygon with $n$ sides be $S_2$. 4. Using the formula: $$S_1 = (n-1 - 2) \times 180 = (n-3) \times 180$$ $$S_2 = (n - 2) \times 180$$ 5. Given the ratio: $$\frac{S_1}{S_2} = \frac{2}{3}$$ 6. Substitute the sums: $$\frac{(n-3) \times 180}{(n-2) \times 180} = \frac{2}{3}$$ 7. Simplify by canceling 180: $$\frac{n-3}{n-2} = \frac{2}{3}$$ 8. Cross multiply: $$3(n-3) = 2(n-2)$$ $$3n - 9 = 2n - 4$$ 9. Solve for $n$: $$3n - 2n = -4 + 9$$ $$n = 5$$ 10. Now find the interior angles of each polygon. 11. For polygon with $n-1 = 4$ sides (a square): $$\text{Interior angle} = \frac{(4-2) \times 180}{4} = \frac{2 \times 180}{4} = 90^\circ$$ 12. For polygon with $n = 5$ sides (a regular pentagon): $$\text{Interior angle} = \frac{(5-2) \times 180}{5} = \frac{3 \times 180}{5} = 108^\circ$$ 13. Final answers: - $n = 5$ - Interior angle of polygon with $4$ sides is $90^\circ$ - Interior angle of polygon with $5$ sides is $108^\circ$