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The problem is asking to find the measure of angle \( \angle AOB \), where \( O \) is the point of intersection of the angle bisectors of angles \( A \) and \( B \) in triangle \( \triangle ABC \). We are given that \( \angle C = 70^\circ \).

### Solution:

In a triangle, the sum of the interior angles is always \( 180^\circ \). So we can find the sum of \( \angle A \) and \( \angle B \) as follows:

\[
\angle A + \angle B + \angle C = 180^\circ
\]

Since \( \angle C = 70^\circ \), we have:

\[
\angle A + \angle B + 70^\circ = 180^\circ
\]

\[
\angle A + \angle B = 110^\circ
\]

Next, because the bisectors of \( \angle A \) and \( \angle B \) intersect at point \( O \), the angle \( \angle AOB \) at the incenter (point \( O \)) is given by the formula:

\[
\angle AOB = 90^\circ + \frac{\angle C}{2}
\]

Substitute \( \angle C = 70^\circ \):

\[
\angle AOB = 90^\circ + \frac{70^\circ}{2}
\]

\[
\angle AOB = 90^\circ + 35^\circ = 125^\circ
\]

### Final Answer:
The measure of \( \angle AOB \) is \( 125^\circ \).

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Would you like further clarification on this solution, or any additional steps explained in detail?

Here are 5 related questions to consider:
1. How do you derive the sum of interior angles of a triangle?
2. What is the importance of the incenter in triangle geometry?
3. Why do we use the formula \( \angle AOB = 90^\circ + \frac{\angle C}{2} \) for this scenario?
4. Can this method be generalized to other types of triangles?
5. How would the solution change if \( \angle C \) were different?

**Tip:** Remember that the incenter of a triangle is equidistant from all sides of the triangle, and is the center of the inscribed circle.

In △ABC, bisectors of ∠A and ∠B intersect at point O. If ∠C = 70°, find the measure of ∠AOB.

This problem focuses on finding the measure of ∠AOB, where angle bisectors intersect at the incenter of △ABC. Learn how to use angle bisectors, incenter properties, and angle sum properties to solve this geometry problem. Suitable for Grades 9-11.

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