2.2 applications vertical angles angle bisector ink.notebook. September 30, page 82. page 81 page 80
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1 page 82 page 81 page apps - Vertical Angles and Angle Bisectors Lesson Objectives Standards page Applications: Vertical Angles, Angle Bisector Press the tabs to view details. 1
2 Standards Lesson Notes Lesson Objectives Standards Lesson Notes N.Q.3 I will use limitations on geometry concepts to understand that angles and lengths cannot be negative A.CED.1 I will create an equation from a word problem and use it to solve for the missing information A.CED.1 I will create an equation from the properties of angle bisectors and use it to solve for the missing angles. A.CED.1 I will create an equation from the properties of vertical angles and use it to solve for the missing angles. A.CED.3 I will understand that when solving equations derived from geometric concepts that the answer cannot have a negative angle or length. A.CED.3 I will demonstrate how an equation with variables on both sides, both with and without distributing can have no solution or an all reals solution A.CED.3 I will demonstrate that distances and angles must be positive when solving geometric concepts A.CED.3 I will identify the constraints on a word problem A.REI.1 I will solve a proportion by using cross multiplication A.REI.1 I will explain each step in solving an equation with variables on both sides A.REI.3 I will solve an equation with variables on both sides with and without distribution A.REI.3 I will explain each step in solving an equation with variables on both sides Press the tabs to view details. A.CED.1 Create equations and inequalities in one variable and use them to solve problems. A.CED.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. N.Q.3 Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. A.REI.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. A.REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. Press the tabs to view details. My Definition: Ray that divides an angle into 2 equal angles Example: Angle Bisector 2x-5 2x - 5 = 30 Characteristics: "=" sign goes in the middle 2x+ x Counterexample: x + 2x = 90 = 180 My Definition: Angles directly across from one another "opposite angles" Example: Vertical Angles 2x x + 4 = 100 Characteristics: "=" sign goes in the middle Counterexample:
3 angle bisector complementary 30 2x-5 2x - 5 = 30 vertical angles 2x+4 = = 100 2x + 4 = 100 midpoint 2x+4 = 28 2x+4 = 28 x + 2x = 90 3x = 90 supplementary 2x + x = 180 3x = 180 triangle angle sum 3x + 2x x 6x = 180 = 180 Equation n = DE = DF = Angle Bisector S V T U 2. 3
4 3. 4. P 40 J Q S M 4x + 5 L K KM is the Û bisector of ÛJKL. R m Û SQR = If m Û JKL = 110, then m Û PQR = m Û JKM = Equation = x = R n - 60 A Q V Y 5n n n Z W D AQ bisects ÛRAD. Equation = Equation = n = n = m Û RAQ = m Û VWY = m Û ZWV = 4
5 Vertical Angles 7. P S 40 R Y V Q T W X Z m Û SQT = m Û ZWX = m Û RQT = m Û XWY = 5
6 9. B 2n + 39 E C D 10. W 6x - 3 5n V X F Equation = Y 5x + 18 U Equation = n = m Û BCD = x = m Û YXU = m Û DCF = m Û ECF = m Û VXU =, m Û WXV = On Your Whiteboards 6
7 a) T B 32 A M AT is the angle bisector of BTM m ATM = b) P R 100 Q 4x S T Equation = c) X W O 3 V Y U x = m RQT = m SQT = d) 1 Find the value of x in the diagram below. Multiple choice A 2 B 4 (6x 2) (2x + 14) C 4 On the Worksheet D 2 7
8 1. 2. Homework Equation Equation Answer 3 Answer 13 Answer 26 Answer 2 Answer 10 Answer Answer 6 Answer 12 Answer 24 Answer 9 Answer 44 Answer 88 8
9 Equation = 9
10 13. Answers: 1) 4x + 1 = 6x - 5, 3, 13, 26 3) 3x - 6 = 2x, 6, 12, 24 5) 5x + 4 = 124, 24, 124, 56 7) 4x + 10 = x + 34, 8, 42, 138 9) 8x - 60 = 180, 30, 30, ) 5x + 10 = 90, 16, 58, 32 13) x + 8 = 2x + 3, 5, 13, 26 10
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