Simplify the complex number (1+2i)/(2+i).

To simplify, multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $2+i$ is $2-i$.

$$\frac{1+2i}{2+i} = \frac{1+2i}{2+i} \times \frac{2-i}{2-i}$$ $$= \frac{(1)(2) + (1)(-i) + (2i)(2) + (2i)(-i)}{(2)(2) + (2)(-i) + (i)(2) + (i)(-i)}$$ $$= \frac{2 - i + 4i - 2i^2}{4 - 2i + 2i - i^2}$$

Substitute $i^2 = -1$:

$$= \frac{2 + 3i - 2(-1)}{4 - (-1)}$$ $$= \frac{2 + 3i + 2}{4 + 1}$$ $$= \frac{4 + 3i}{5}$$ $$= \frac{4}{5} + \frac{3}{5}i$$

The complex number is in the form $x + yi$, where $x = \frac{4}{5}$ and $y = \frac{3}{5}$.
The modulus $r$ is given by the formula $r = \sqrt{x^2 + y^2}$.

$$r = \sqrt{\left(\frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2}$$ $$r = \sqrt{\frac{16}{25} + \frac{9}{25}}$$ $$r = \sqrt{\frac{16+9}{25}}$$ $$r = \sqrt{\frac{25}{25}}$$ $$r = \sqrt{1}$$ $$r = 1$$

The problem states that $\frac{1+2i}{2+i} = r(\cos \theta + i \sin \theta)$.
From our calculation, we found that $\frac{1+2i}{2+i} = 1(\cos \theta + i \sin \theta)$.
By comparing the two expressions, we can see that $r = 1$.

The final answer is $\text{c. } 1$.