The image contains three geometry problems. The first problem (labeled 10) asks to find angles \angle 2, \angle 3, and \angle 4 given that \angle 1 = 125 degrees. The second problem (labeled 12) asks to find the sum of angles \angle 1 + \angle 2 + \angle 3. The third problem (labeled 13) asks to find the measure of \angle AOC.

Problem 10:

We are given that \(∠1 = 125^°). Lines a and b are perpendicular, so \(∠3 + ∠4 = 90^°) and \(∠1 + ∠2 = 180^°) (linear pair).

1. Find ∠2: Since \(∠1 + ∠2 = 180^°), then \(∠2 = 180^° - 125^° = 55^°).
2. Find ∠3: Since lines a and b are perpendicular, \(∠3 + ∠4 = 90^°). Also, \(∠2 + ∠3 = 90^°) (vertical angles are equal, and the angle formed by the intersection of lines a and b is 90 degrees). Thus, \(∠3 = 90^° - 55^° = 35^°).
3. Find ∠4: Since \(∠3 + ∠4 = 90^°), then \(∠4 = 90^° - 35^° = 55^°).

Answer for Problem 10: \(∠2 = 55^°, ∠3 = 35^°, ∠4 = 55^°).

Problem 12:

Angles \(∠1, ∠2, and ∠3) form a straight line. Therefore, they are supplementary.

Answer for Problem 12: \(∠1 + ∠2 + ∠3 = 180^°).

Problem 13:

We are given the measures of \(∠DOF = 30^°) and \(∠BOE = 40^°). Angles \(∠AOD, ∠BOE, ∠COF) are vertically opposite to \(∠BOC, ∠AOE, ∠DOF) respectively. Angles on a straight line add up to 180 degrees.

1. Find ∠AOD: \(∠AOD) and \(∠BOC) are vertically opposite, so \(∠AOD = ∠BOC). Angles \(∠COE, ∠EOB, ∠BOF, ∠FOA) form a full circle (360 degrees). Also, \(∠COE) and \(∠DOF) are vertically opposite, so \(∠COE = ∠DOF = 30^°). Similarly, \(∠AOE) and \(∠BOF) are vertically opposite.
2. Find ∠AOC: \(∠AOC) is formed by \(∠AOF) + \(∠FOC). We know \(∠DOF = 30^°) and \(∠BOE = 40^°). Since \(∠COE = ∠DOF = 30^°) and \(∠AOE + ∠EOB = ∠AOB). Line AE is a straight line, so \(∠AOC + ∠COE + ∠EOB = 180^°). Let's consider the straight line CE. Angles on this line are \(∠COF, ∠FOD, ∠DOE). However, this does not seem to be a straight line. Let's consider line AE as a straight line. Then \(∠AOC + ∠COE + ∠EOB = 180^°). We know \(∠COE = 30^°) (vertically opposite to \(∠DOF)). We know \(∠BOE = 40^°). So, \(∠AOC + 30^° + 40^° = 180^°). Therefore, \(∠AOC = 180^° - 70^° = 110^°).

Answer for Problem 13: \(∠AOC = 110^°).