We draw a triangle inside of a circle with one vertex at the center of the circle and the other two vertices on the circumference of the circle. The angle at the center of the circle measures $75$ degrees. We draw a second triangle, congruent to the first, also with one vertex at the center of the circle and the other vertices on the circumference of the circle rotated $75$ degrees clockwise from the first triangle so that it shares a side with the first triangle. We draw a third, fourth, and fifth such triangle each rotated $75$ degrees clockwise from the previous triangle. The base of the fifth triangle will intersect the base of the first triangle. What is the degree measure of the obtuse angle formed by the intersection?
1. Let the center of the circle be \( O \), and let \( A_1 \) and \( A_2 \) be the points on the circumference of \( O \) such that \( \angle A_1 O A_2 = 75^\circ \). Continue this labeling around the circle, until we reach \( A_6 \).

2. Since each triangle is rotated \( 75^\circ \) clockwise from the previous one, the angle between \( A_1 \) and \( A_6 \) at the center \( O \) is:
   \[
   75^\circ \times 5 = 375^\circ
   \]
   However, since a full circle is \( 360^\circ \), we subtract \( 360^\circ \) to find the actual angle:
   \[
   375^\circ - 360^\circ = 15^\circ
   \]
   Therefore, \( \angle A_1 O A_6 = 15^\circ \).

3. Each triangle is isosceles with the two equal sides being the radii of the circle. Therefore, the base angles of each triangle are:
   \[
   \angle OA_1A_2 = \angle OA_2A_1 = \frac{180^\circ - 75^\circ}{2} = \frac{105^\circ}{2} = 52.5^\circ
   \]
   Similarly, for the other triangles:
   \[
   \angle OA_2A_3 = \angle OA_3A_2 = 52.5^\circ
   \]
   and so on up to:
   \[
   \angle OA_6A_5 = \angle OA_5A_6 = 52.5^\circ
   \]

4. Consider the quadrilateral \( A_1 X A_6 O \), where \( X \) is the intersection of \( A_1A_2 \) and \( A_6A_5 \). The sum of the internal angles of a quadrilateral is \( 360^\circ \). We know:
   \[
   \angle OA_1A_2 + \angle OA_6A_5 + \angle A_1OA_6 + \measuredangle A_6XA_1 = 360^\circ
   \]
   Substituting the known angles:
   \[
   52.5^\circ + 52.5^\circ + 15^\circ + \measuredangle A_6XA_1 = 360^\circ
   \]
   Simplifying:
   \[
   120^\circ + \measuredangle A_6XA_1 = 360^\circ
   \]
   Therefore:
   \[
   \measuredangle A_6XA_1 = 360^\circ - 120^\circ = 240^\circ
   \]

5. The reflex angle \( \measuredangle A_6XA_1 = 240^\circ \) means the internal angle \( \angle A_1XA_6 \) is:
   \[
   360^\circ - 240^\circ = 120^\circ
   \]

The final answer is \( \boxed{120^\circ} \).