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Let's solve this problem step-by-step.

1. **Identify the given information:**
   - \(\angle ACD = 105^\circ\)
   - \(\angle EAF = 45^\circ\)
   
   These are external angles of the triangle.

2. **Calculate interior angles using the external angle properties:**

   - **For \(\angle ABC\):**
     Since \(\angle ACD\) is an external angle to \(\triangle ABC\) at vertex \(C\), it will be equal to the sum of the opposite interior angles:
     \[
     \angle ABC + \angle BAC = \angle ACD = 105^\circ
     \]
   
   - **For \(\angle BAC\):**
     Similarly, \(\angle EAF\) is an external angle to \(\triangle ABC\) at vertex \(A\). So:
     \[
     \angle BAC + \angle BCA = \angle EAF = 45^\circ
     \]

3. **Set up equations:**

   Let:
   - \(\angle ABC = x\)
   - \(\angle BAC = y\)
   - \(\angle BCA = z\)

   We now have two equations from the given external angles:
   - \(x + y = 105^\circ\)
   - \(y + z = 45^\circ\)

4. **Use the triangle angle sum property:**

   For any triangle, the sum of its internal angles is \(180^\circ\):
   \[
   x + y + z = 180^\circ
   \]

5. **Solve the equations:**

   Substitute \(y\) from the first equation into the third equation to solve for the angles:
   - From \(x + y = 105^\circ\), express \(y\) as \(y = 105^\circ - x\).
   - Substitute \(y\) in the second equation:
     \[
     (105^\circ - x) + z = 45^\circ
     \]
     Simplifying, we get:
     \[
     z = x - 60^\circ
     \]
   So that

In the following figure, the sides BC, CA, and BA of a triangle ABC have been produced to points D, E, and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the triangle ABC.

Calculate the interior angles of triangle ABC when given the exterior angles ∠ACD = 105° and ∠EAF = 45°. This problem involves the Exterior Angle Theorem and is suitable for students in Grades 8-10.

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