Lines And Angles

100°

Step 1: Let the angles be 4x and 5x since they are in ratio 4:5. Step 2: Since they are supplementary, their sum equals 180°: 4x + 5x = 180°. Step 3: Simplify: 9x = 180°, so x = 20°. Step 4: The larger angle = 5x = 5(20°) = 100°. Step 5: Verify: 80° + 100° = 180° ✓

70°

Step 1: Use the exterior angle theorem - an exterior angle equals the sum of the two non-adjacent interior angles. Step 2: The exterior angle at Q = ∠P + ∠R. Step 3: Substitute: 120° = 50° + ∠R. Step 4: Solve: ∠R = 120° - 50° = 70°. Step 5: Verify using angle sum: ∠P + ∠Q + ∠R = 50° + 60° + 70° = 180° ✓

105°

Step 1: Identify that we have co-interior angles (interior angles on the same side of the transversal). Step 2: Recall the property that co-interior angles are supplementary when lines are parallel. Step 3: Set up equation: 75° + other angle = 180°. Step 4: Solve: other angle = 180° - 75° = 105°. Step 5: This confirms the co-interior angle property.

70°

Step 1: In isosceles triangle ABC with AB = AC, the base angles ∠B and ∠C are equal. Step 2: Apply angle sum property: ∠A + ∠B + ∠C = 180°. Step 3: Since ∠B = ∠C, we have: 40° + ∠B + ∠B = 180°. Step 4: Simplify: 40° + 2∠B = 180°. Step 5: Solve: 2∠B = 140°, so ∠B = 70°. Therefore, each base angle is 70°.