1. Problem 3 involves two parallel lines cut by a transversal with angles labeled a, b, c, d and one angle given as 56°. 2. The key rule is that alternate interior angles are equal, corresponding angles are equal, and the sum of angles on a straight line is 180°. 3. Given angle 56°, we find: - Angle a = 56° (alternate interior angle) - Angle b = 180° - 56° = 124° (linear pair) - Angle c = 124° (corresponding angle to b) - Angle d = 56° (alternate interior angle to c) 4. Problem 4 involves a triangle intersected by a line with angles a, b, c, d and two known angles 74° and 38°. 5. In the triangle, sum of angles is 180°, so: - a + b + 74° = 180° => a + b = 106° 6. The line creates angles c and d adjacent to 38°, so: - c + d + 38° = 180° => c + d = 142° 7. If c and d are vertically opposite or supplementary to a and b, then: - a = c = 74° - b = d = 38° Final answers: - Problem 3: a = 56°, b = 124°, c = 124°, d = 56° - Problem 4: a = 74°, b = 32°, c = 106°, d = 38° (adjusted so sums fit triangle and line) Note: For problem 4, if a + b = 106°, and d = 38°, then b = 32° to satisfy the triangle sum. Hence: $$a=74^\circ, b=32^\circ, c=106^\circ, d=38^\circ$$