Special angle pairs can help you identify geometric relationships. We use them to find angle measures. 1.5 Exploring Angle Pairs Problem 1: Use the diagram at the right. Is the statement true? Explain a.
Problem 2: What can you conclude from the information in the diagram? 1.5 Exploring Angle Pairs Problem 2 Solution: < 1 and < 2 are congruent. This means they have the same measurement in degrees. < 1 + < 2 are congruent to < 4.
This means they have the same measurement in degrees. They are also opposites which means they are vertical angles. < 3 and < 5 are vertical and congruent. 1.5 Exploring Angle Pairs Problem 2b: Can you make each conclusion from the information in the diagram? Explain.
a. b. c. d. Segment TW is congruent to Segment WV Segment PW is congruent to Segment WQ
Problem 2b Solutions: Can you make each conclusion from the information in the diagram? Explain. a. Segment TW is congruent to Segment WV Since they have a red mark , YES they are congruent. b. Segment PW is congruent to Segment WQ We cannot conclude this. No marks are given and we do not know the length. c.
d. Segment TV bisects Segment PQ No. Segment PQ is bisecting segment TV. 1.5 Exploring Angle Pairs A linear pair is a pair of adjacent angles whose noncommon sides are opposite rays. The angles of a linear pair form a straight angle. Remember a straight line = 1800 So supplementary angles = 1800 1.5 Exploring Angle Pairs
1. Make a sketch (drawing) 2. Label 2x + 24 3. Use algebra to solve. (2x + 24) + (4x + 36) = 180 6x + 60 = 180 6x = 120 x = 20 K
L 4x + 36 P J 4. Plug in the variable and find the measure of the angle. < KPL = (2(20) + 24) = 40 + 24 = 64 < JPL = (4(20) + 36) = 80 + 36 = 116 1.5 Exploring Angle Pairs
An angle bisector is a ray that divides an angle into two congruent (same degree) angles. Its endpoint is a the angle vertex. Within the ray, a segment with the same endpoint is also an angle bisector. The ray or segment bisects the angle. In the diagram, Ray AY is the angle bisector of
Ray AC bisects
58 D 3. Use algebra to solve. Since AC bisects (cuts in half) ADB, we know that
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