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In this problem, we are given:

- \( \angle CAP = 34^\circ \)
- \( S \) is the incenter of \( \triangle ABC \)

Since \( S \) is the incenter, it is the point where the angle bisectors of \( \triangle ABC \) meet. Therefore:

1. \( \angle CAP \) is half of \( \angle CAB \) because \( AP \) is an angle bisector of \( \angle CAB \).

To find \( \angle CAB \):
\[
\angle CAB = 2 \times \angle CAP = 2 \times 34^\circ = 68^\circ
\]

So, the correct answer is:
\[
\angle CAB = 68^\circ
\]

Would you like further details or have any questions?

Here are some additional related questions for further exploration:

1. How can the incenter property be used to find other angle measures in a triangle?
2. If the incenter divides another angle in the triangle, what relationships exist between the divided angles?
3. How does knowing one angle bisector affect understanding of a triangle’s properties?
4. What properties define an incenter and differentiate it from other triangle centers, like the centroid?
5. How would the solution differ if \( S \) were the centroid instead of the incenter?

**Tip:** Remember, the incenter is equidistant from each side of the triangle and is the point of intersection for all angle bisectors.

Learn how to find ∠CAB in triangle ABC, where S is the incenter and m∠CAP = 34°. This solution involves geometry concepts like angle bisectors and the incenter theorem, making it suitable for students in Grades 8-10.

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