A strip is defined as the region between two parallel lines; the width of the strip is the distance between the two lines. Two strips of width $1$ intersect in a parallelogram whose area is $2.$ What is the angle between the strips?
\[
\mathrm a. ~ 15^\circ\qquad \mathrm b.~30^\circ \qquad \mathrm c. ~45^\circ \qquad \mathrm d. ~60^\circ \qquad \mathrm e. ~90^\circ\]
1. Consider two strips of width 1 that intersect to form a parallelogram \(ABCD\) with an area of 2. The width of each strip is the perpendicular distance between the parallel lines defining the strip.
2. Let the parallel lines defining the first strip be \(a\) and \(b\), and the parallel lines defining the second strip be \(f\) and \(g\). The distance between \(a\) and \(b\) is 1, and the distance between \(f\) and \(g\) is also 1.
3. The area of the parallelogram \(ABCD\) is given by the formula:
   \[
   \text{Area} = \text{base} \times \text{height}
   \]
   Here, the base is the length of one side of the parallelogram, and the height is the perpendicular distance between the opposite sides.
4. Let \(\overline{AB}\) be the base of the parallelogram. Since the area of the parallelogram is 2 and the height (distance between the parallel lines) is 1, we have:
   \[
   \text{Area} = \overline{AB} \times 1 = 2 \implies \overline{AB} = 2
   \]
5. Let \(P\) be the foot of the perpendicular from \(A\) to \(BC\). Since \(\overline{AP} = 1\) (the width of the strip), \(\triangle APB\) is a right triangle with \(\overline{AP} = 1\) and \(\overline{AB} = 2\).
6. In \(\triangle APB\), we can use trigonometric relationships to find the angle \(\angle PBA\). Since \(\overline{AP} = 1\) and \(\overline{AB} = 2\), we have:
   \[
   \cos(\angle PBA) = \frac{\overline{AP}}{\overline{AB}} = \frac{1}{2}
   \]
7. The angle whose cosine is \(\frac{1}{2}\) is \(60^\circ\). Therefore, \(\angle PBA = 60^\circ\).
8. However, since \(\angle PBA\) is the angle between the strips, we need to consider the angle between the lines defining the strips. The angle between the strips is the complement of \(\angle PBA\), which is:
   \[
   90^\circ - 60^\circ = 30^\circ
   \]

The final answer is \(\boxed{30^\circ}\)