Module 10: Circles Topic 2 Application: Central and Inscribed Angles Problem Set 3.

Step 1 ：Observe the triangle with side lengths 2, 3, and 4. The angle opposite the side of length 2 is \(\frac{\alpha}{2}\).

Step 2 ：Apply the Law of Cosines to this triangle: \(2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos\frac{\alpha}{2}\).

Step 3 ：Rearrange the equation to solve for \(\cos\frac{\alpha}{2}\): \(21 = 24\cos\frac{\alpha}{2}\) which simplifies to \(\cos\frac{\alpha}{2} = \frac{7}{8}\).

Step 4 ：Use the double angle formula \(\cos 2\theta = 2\cos^2 \theta - 1\) to find \(\cos\alpha\). Substituting \(\frac{7}{8}\) for \(\cos\frac{\alpha}{2}\) gives \(\cos\alpha = \frac{17}{32}\).

Step 5 ：The sum of the numerator and denominator of \(\cos\alpha\) is 17 + 32 = 49.

Step 6 ：The final answer is \(\boxed{49}\).