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Chapter 1 Section 7

Chapter 1 Section 7. Angle Relationships. Warm-Up. Refer to the figure below. B. 1. G. 2. 3. D. C. A. Which angles have D as their vertex? 2) Give two other names for <ABD. 3) Does <BDC appear to be acute, right, obtuse, or straight?

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Chapter 1 Section 7

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  1. Chapter 1Section 7 Angle Relationships

  2. Warm-Up • Refer to the figure below. B 1 G 2 3 D C A Which angles have D as their vertex? 2) Give two other names for <ABD. 3) Does <BDC appear to be acute, right, obtuse, or straight? 4) Find the value of x and m<ABD if m<ABC = 71, m<DBC = 2x – 3, and m<ABD = 3x + 4. 5) Suppose MN bisects <RMQ, m<RMN = x + 2, and m<NMQ = 2x – 3. Find the value of x and m<RMQ.

  3. Warm-Up • Refer to the figure below. B 1 G 2 3 D C A Which angles have D as their vertex? <ADB and <BDC 2) Give two other names for <ABD. <DBA and <1 3) Does <BDC appear to be acute, right, obtuse, or straight? Obtuse

  4. Warm-Up • Refer to the figure below. B 1 G 2 3 D C A 4) Find the value of x and m<ABD if m<ABC = 71, m<DBC = 2x – 3, and m<ABD = 3x + 4. Use the angle addition postulate. m<ABD + m<DBC = m<ABC 3x + 4 + 2x – 3 = 71 5x + 1 = 71 5x = 70 x = 14 Plug 14 in for x in the equation for m<ABD. m<ABD = 3x + 4 m<ABD = 3(14) + 4 m<ABD = 42 + 4 m<ABD = 46

  5. 5) Suppose MN bisects <RMQ, m<RMN = x + 2, and m<NMQ = 2x – 3. Find the value of x and m<RMQ. R N M Q Since MN bisects <RMQ, we know that m<RMN = m<NMQ. m<RMN = m<NMQ x + 2 = 2x – 3 2 = x – 3 5 = x Plug x in to either equation and multiple by two(angle addition postulate). m<RMQ = 2(m<RMN) m<RMQ = 2(x + 2) m<RMQ = 2x + 4 m<RMQ = 2(5) + 4 m<RMQ = 10 + 4 m<RMQ = 14

  6. Vocabulary Perpendicular lines- Lines that intersect to form right angles. Adjacent Angles- Angles in the same plane that have a common vertex and a common side, but no common interior points. (<1 and <2 are adjacent angles) Vertical Angles- Two nonadjacent angles formed by two intersecting lines. (<3 and <4 are vertical angles) Vertical angles are congruent. 1 2 3 4

  7. Vocabulary Cont. Linear Pair- Adjacent angles whose noncommon sides are opposite rays.( <1 and <2 form a linear pair) Supplementary angles- Two angles whose measures have a sum of 180. (<3 and <4 are supplementary) Complementary angles- Two angles whose measures have a sum of 90. (<5 and <6 are complementary) 1 2 3 4 5 6

  8. Example 1: Use the figure below to find the value of x and m<JIH. J G 16x - 20 I H K 13x + 7 Since <GIJ and <KIH are vertical angles, we know they are congruent. m<GIJ = m<KIH 16x – 20 = 13x + 7 3x – 20 = 7 3x = 27 x = 9 Since <KIH and <JIH are supplementary, we know they add up to 180. Plug 9 in for x in the equation for <KIH. m<KIH + m<JIH = 180 13x + 7 + m<JIH = 180 13(9) + 7 + m<JIH = 180 117 + 7 + m<JIH = 180 124 + m<JIH = 180 m<JIH = 56

  9. Example 2: Suppose the measure of angle GIJ = 9x – 4 and the measure of angle JIH = 4x – 11. Find the value of x and the measure of angle KIH. J G 9x - 4 4x - 11 I H K Since <GIJ and <JIH are supplementary and form a linear pair, their measures add up to 180. m<GIJ + m<JIH = 180 9x – 4 + 4x – 11 = 180 13x – 15 = 180 13x = 195 x = 15 Since <KIH and <GIJ are vertical angles, they are congruent or have equal measures. Plug 15 in for x in the equation for m<GIJ. m<GIJ = 9x – 4 m<GIJ = 9(15) - 4 m<GIJ = 135 – 4 m<GIJ = 131 = m<KIH

  10. Example 3: The measure of the complement of an angle is 3.5 times smaller than the measure of the supplement of the angle. Find the measure of the angle. Let x be the measure of the angle. Then 180 – x is the measure of its supplement, and 90 – x is the measure of its complement. 180 – x = 3.5(90 – x) 180 – x = 3.5(90 – x) 180 – x = 315 – 3.5x 180 + 2.5x = 315 2.5x = 135 x = 54 So the measure of the angle is 54 degrees. Supplement is Complement 3.5 times smaller then supplement

  11. Example 4: The measure of the supplement of an angle is 60 less than three times the measure of the complement of the angle. Find the measure of the angle. Let x be the measure of the angle. Then 180 – x is the measure of its supplement, and 90 – x is the measure of its complement. 180 – x = 3(90 – x) – 60 180 – x = 3(90 – x) – 60 180 – x = 270 – 3x – 60 180 – x = 210 – 3x 180 + 2x = 210 2x = 30 x = 15 So the measure of the angle is 15 degrees. Supplement is 60 less than three times the complement

  12. Example 5: Find m<S if m<S is 20 more than four times its supplement. Let x = m<S. Then 180 – x is the measure of its supplement. x = 4(180 – x) + 20 x = 4(180 – x) + 20 x = 720 - 4x + 20 x = 740 – 4x 5x = 740 x = 148 So m<S is 148 degrees. m<S is 20 more than four times its supplement

  13. Example 6: Refer to the figure at the right. C D B F E A Identify a pair of obtuse vertical angles. <AFE and <BFD B) Name a segment that is perpendicular to ray FC. Ray AD C) Which angle forms a linear pair with angle DFE? <EFA or <DFB D) Which angle is complementary to <CFB? <BFA

  14. Example 7: <N is a complement of <M, m<N = 8x – 6 and m<M = 14x + 8. Find the value of x and m<M. Since the two angles are complements, we know they add up to 90. m<N + m<M = 90 8x – 6 + 14x + 8 = 90 22x + 2 = 90 22x = 88 x = 4 Plug 4 in for x in the equation for m<M. m<M = 14x + 8 m<M = 14(4) + 8 m<M = 56 + 8 m<M = 65

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