Mrs. Rivas. (x − 26) + x = 180 x − 26 + x = 180 2x − 26 = 180 2x = 206 x = 103 103 77 Same-side Interior angles 103 + 77 = 180 Mrs. Rivas.

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Presentation transcript:

Mrs. Rivas

(x − 26) + x = 180 x − 26 + x = 180 2x − 26 = 180 2x = 206 x = Same-side Interior angles = 180 Mrs. Rivas

Corresponding angles (3x − 5) = (x + 55) 3x − 5 = x x − 5 = 55 2x = 60 x = 30 3(30) − 5 = − 5 = = 85 Mrs. Rivas

(x + 20) + (x + 10) = 180 Linear Pair angles = Suppl. x x + 10 = 180 2x + 30 = 180 2x = 150 x = = 180 Mrs. Rivas

(y − 40) + y = 180 Linear Pair angles = Suppl. y − 40 + y = 180 2y − 40 = 180 2x = 220 x = = 180 Mrs. Rivas

Alternate Exterior angles (2x + 6) = 42 2x + 6 = 42 2x = 36 x = Mrs. Rivas

(3x − 17) + 98 = 180 3x − = 180 3x + 81 = 180 3x = 99 x = 33 Same-side Interior angles = 180 Mrs. Rivas

2x x + 24 = 180 8x + 44 = 180 8x = 136 x = 17 Same-side Interior angles (2x + 20) + (6x + 24) = = 180 Mrs. Rivas

Alternate Exterior angles (2x + 2) = (3x − 63) 2x + 2 = 3x − 63 2x = 3x − 65 − x = − 65 x = Mrs. Rivas

Linear Pair angles = Suppl. (5x − 5) = 180 5x − = 180 5x = 180 5x = 45 x = = 180 Mrs. Rivas

Same-side Interior angles (4x − 8) = 180 4x − = 180 4x = 180 4x = 76 x = = 180 Mrs. Rivas

(7x + 14) + (2x + 4) = 180 7x x + 4 = 180 9x + 18 = 180 9x = 162 x = = 180 Linear Pair angles = Suppl. Mrs. Rivas

(4x − 5) + (x + 20) = 180 4x − 5 + x + 20 = 180 5x + 15 = 180 5x = 165 x = = 180 Linear Pair angles = Suppl. Mrs. Rivas

19. Developing Proof Complete the flow proof below. Given:  1 and  4 are supplementary. Given Vertical ∠s are ≅ ∠2 ∠3 are supplementary ∠2 and ∠3 are supplementary Converse of same-side Interior Angles

Mrs. Rivas 20. Developing Proof Complete the flow proof below. ∠2 ∠3 are supplementary ∠2 and ∠3 are supplementary Give ∠s ∠s suppl. to the same ∠ are ≅ ∠1 ∠4 ∠1 ≅ ∠4