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OA bisects <COL. <COA = 3x + 3, <AOL = 5x – 13 Draw a picture to represent this situation.

C. A. 3 x + 3 . OA bisects <COL. <COA = 3x + 3, <AOL = 5x – 13 Draw a picture to represent this situation. . 5x – 13 . O. L. 3x + 3 = 5x – 13 16 = 2x 8 = x. Now set up an equation and solve for x. 10x – 8 . 3x + 6 . 1. 2. 10x – 8 + 3x + 6 = 180

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OA bisects <COL. <COA = 3x + 3, <AOL = 5x – 13 Draw a picture to represent this situation.

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  1. C A 3x + 3 OA bisects <COL. <COA = 3x + 3, <AOL = 5x – 13 Draw a picture to represent this situation. 5x – 13 O L 3x + 3 = 5x – 13 16 = 2x 8 = x Now set up an equation and solve for x.

  2. 10x – 8 3x + 6 1 2 10x – 8 + 3x + 6 = 180 13x – 2 = 180 13x = 182 x = 14 Find the measure of angle 1. <1 = 10(14) – 2 = 138˚

  3. 8x + 6 = 14y 8x – 14y = -6 Q 20x – 4y Z 8x + 6 8x + 6 + 14y = 20x – 4y -12x + 18y = -6 UZ bisects <QUI. <QUZ = 8x + 6 <ZUI = 14y<QUI = 20x – 4yDraw a picture to represent this situation. 14y U I 2(14y) = 20x – 4y 28y = 20x – 4y -20x + 32y = 0 Now set up a system of equations and solve for x and y. 2(8x + 6) = 20x – 4y 16x + 12 = 20x – 4y -4x + 4y = -12 x = 8 y = 5

  4. 3x = 16 + 180 – x 4x = 196 x = 49 Complement = 90 – 49 = 41˚ Three times an angle is 16 more than the angle’s supplement. Find the measure of the angle’s complement.

  5. 4x + 3 2x + 53 A T C A is the midpoint of segment CT.CA = 4x + 3AT = 2x + 53 Draw a picture to represent this situation. Now set up an equation and solve for x. 4x + 3 = 2x + 53 2x = 50 x = 25

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