During chemistry labs, we oftentimes fold a disk-shaped filter paper twice, and then open up a flap of the quartercircle to form a cone shape, as in the diagram. What is the angle $\theta$, in degrees, of the bottom of the cone when we look at it from the side?
[img]https://cdn.artofproblemsolving.com/attachments/d/2/f8e3a7afb606dfd6fad277f547b116566a4a91.png[/img]
1. **Understanding the problem**: We start with a disk-shaped filter paper of radius \( r \). When folded twice, it forms a quarter circle. When one flap is opened, it forms a cone. We need to find the angle \(\theta\) at the bottom of the cone when viewed from the side.

2. **Forming the cone**: When the quarter circle is opened to form a cone, the arc length of the quarter circle becomes the circumference of the base of the cone. The arc length of the quarter circle is:
   \[
   \text{Arc length} = \frac{1}{4} \times 2\pi r = \frac{\pi r}{2}
   \]

3. **Base circumference of the cone**: Let the radius of the base of the cone be \( R \). The circumference of the base of the cone is:
   \[
   2\pi R = \frac{\pi r}{2}
   \]
   Solving for \( R \):
   \[
   R = \frac{r}{4}
   \]

4. **Slant height of the cone**: The slant height of the cone is the radius of the original disk, which is \( r \).

5. **Height of the cone**: Using the Pythagorean theorem in the right triangle formed by the radius \( R \), the height \( h \), and the slant height \( r \):
   \[
   r^2 = R^2 + h^2
   \]
   Substituting \( R = \frac{r}{4} \):
   \[
   r^2 = \left(\frac{r}{4}\right)^2 + h^2
   \]
   \[
   r^2 = \frac{r^2}{16} + h^2
   \]
   \[
   h^2 = r^2 - \frac{r^2}{16}
   \]
   \[
   h^2 = \frac{16r^2 - r^2}{16}
   \]
   \[
   h^2 = \frac{15r^2}{16}
   \]
   \[
   h = \frac{r\sqrt{15}}{4}
   \]

6. **Angle \(\theta\)**: The angle \(\theta\) at the bottom of the cone is the angle between the two slant heights when viewed from the side. This angle can be found using the properties of the 30-60-90 triangle formed by the radius, height, and slant height of the cone. In a 30-60-90 triangle, the angles are 30°, 60°, and 90°. The angle \(\theta\) is the 60° angle.

Therefore, the angle \(\theta\) is:
\[
\theta = 60^\circ
\]

The final answer is \(\boxed{60^\circ}\).