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 $125^\circ$.

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.