Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM 7•6 Lesson 2: Solving for Unknown Angles Using Equations Student Outcomes ο§ Students solve for unknown angles in word problems and in diagrams involving complementary, supplementary, vertical, and adjacent angles. Classwork Opening Exercise (5 minutes) Opening Exercise Two lines meet at a point. In a complete sentence, describe the relevant angle relationships in the diagram. Find the values of π, π, and π. MP.6 The two intersecting lines form two pairs of vertical angles; π = ππ, and π° = π°. Angles π° and π° are angles on a line and sum to πππ°. r° s° 25° t° π = ππ Scaffolding: π + ππ = πππ π + ππ − ππ = πππ − ππ π = πππ, π = πππ In the following examples and exercises, students set up and solve an equation for the unknown angle based on the relevant angle relationships in the diagram. Model the good habit of always stating the geometric reason when you use one. This is a requirement in high school geometry. Example 1 (4 minutes) Example 1 Two lines meet at a point that is also the endpoint of a ray. In a complete sentence, describe the relevant angle relationships in the diagram. Set up and solve an equation to find the value of π and π. Students may benefit from repeated practice drawing angle diagrams from verbal descriptions. For example, tell them “Draw a diagram of two supplementary angles, where one has a measure of 37°.” Students struggling to organize their solution to a problem may benefit from the five-part process of the Exit Ticket in Lesson 1, including writing an equation, explaining the connection between the equation and the situation, and assessing whether an answer is reasonable. This builds conceptual understanding. The angle π° is vertically opposite from and equal to the sum of the angles with measurements ππ° and ππ°, or a sum of ππ°. Angles π° and π° are angles on a line and sum to πππ°. r° π = ππ + ππ π = ππ Vert. ∠s π + (ππ) = πππ π + ππ − ππ = πππ − ππ π = πππ Lesson 2: ∠s on a line Solving for Unknown Angles Using Equations This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M6-TE-1.3.0-10.2015 28° p° 16° 22 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM 7•6 Take the opportunity to distinguish the correct usage of supplementary versus angles on a line in this example. Remind students that supplementary should be used in reference to two angles, whereas angles on a line can be used for two or more angles. Exercise 1 (4 minutes) Exercise 1 Three lines meet at a point. In a complete sentence, describe the relevant angle relationship in the diagram. Set up and solve an equation to find the value of π. The two π° angles and the angle πππ° are angles on a line and sum to πππ°. ππ + πππ = πππ ∠s on a line ππ + πππ − πππ = πππ − πππ ππ = ππ π = ππ Example 2 (4 minutes) Encourage students to label diagrams as needed to facilitate their solutions. In this example, the label π¦° is added to the diagram to show the relationship of π§° with 19°. This addition allows for methodical progress toward the solution. Example 2 Three lines meet at a point. In a complete sentence, describe the relevant angle relationships in the diagram. Set up and solve an equation to find the value of π. Let π° be the angle vertically opposite and equal in measurement to ππ°. The angles π° and π° are complementary and sum to ππ°. π + π = ππ π + ππ = ππ z° Complementary ∠s πΛ 19° π + ππ − ππ = ππ − ππ π = ππ Lesson 2: Solving for Unknown Angles Using Equations This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M6-TE-1.3.0-10.2015 23 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM 7•6 Exercise 2 (4 minutes) Exercise 2 C Three lines meet at a point; ∠π¨πΆπ = πππ°. In a complete sentence, describe the relevant angle relationships in the diagram. Set up and solve an equation to determine the value of π. ∠π¬πΆπ©, formed by adjacent angles ∠π¬πΆπͺ and ∠πͺπΆπ©, is vertical to and equal in measurement to ∠π¨πΆπ. The measurement of ∠π¬πΆπ© is π° + ππ° (∠s add). π + ππ = πππ E c° O A B 144° Vert. ∠s F π + ππ − ππ = πππ − ππ D π = ππ Example 3 (4 minutes) Example 3 Two lines meet at a point that is also the endpoint of a ray. The ray is perpendicular to one of the lines as shown. In a complete sentence, describe the relevant angle relationships in the diagram. Set up and solve an equation to find the value of π. The measurement of the angle formed by adjacent angles of ππ° and ππ° is the sum of the adjacent angles. This angle is vertically opposite and equal in measurement to the angle π°. Let π° be the measure of the indicated angle. π = πππ π = (π) ∠s add Vert. ∠s 26° πΛ Λ t° π = πππ Exercise 3 (4 minutes) Exercise 3 Two lines meet at a point that is also the endpoint of a ray. The ray is perpendicular to one of the lines as shown. In a complete sentence, describe the relevant angle relationships in the diagram. You may add labels to the diagram to help with your description of the angle relationship. Set up and solve an equation to find the value of π. v° πΛ 46° One possible response: Let π° be the angle vertically opposite and equal in measurement to ππ°. The angles π° and π° are adjacent angles, and the angle they form together is equal to the sum of their measurements. π = ππ π = ππ + ππ Vert. ∠s ∠s add π = πππ Lesson 2: Solving for Unknown Angles Using Equations This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M6-TE-1.3.0-10.2015 24 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM 7•6 Example 4 (4 minutes) Example 4 Three lines meet at a point. In a complete sentence, describe the relevant angle relationships in the diagram. Set up and solve an equation to find the value of π. Is your answer reasonable? Explain how you know. The angle π° is vertically opposite from the angle formed by the right angle that contains and shares a common side with an π° angle. π = ππ − π ∠s add and vert. ∠s x° π = ππ The answer is reasonable because the angle marked by π° is close to appearing as a right angle. 8° Exercise 4 (4 minutes) Exercise 4 Two lines meet at a point that is also the endpoint of two rays. In a complete sentence, describe the relevant angle relationships in the diagram. Set up and solve an equation to find the value of π. Find the measurements of ∠π¨πΆπ© and ∠π©πΆπͺ. ∠π¨πΆπͺ is vertically opposite from the angle formed by adjacent angles ππ° and ππ°. ππ + ππ = ππ + ππ ∠s add and vert. ∠s ππ = πππ 25° A π = ππ 2x° ∠π¨πΆπͺ = π(ππ)° = ππ° O 3x° B ∠π©πΆπͺ = π(ππ)° = ππ° C Lesson 2: Solving for Unknown Angles Using Equations This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M6-TE-1.3.0-10.2015 25 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM 7•6 Exercise 5 (4 minutes) Exercise 5 a. In a complete sentence, describe the relevant angle relationships in the diagram. Set up and solve an equation to find the value of π. Find the measurements of ∠π¨πΆπ© and ∠π©πΆπͺ. ∠π¨πΆπ© and ∠π©πΆπͺ are complementary and sum to ππ°. ππ + (ππ + ππ) = ππ O A complementary ∠s 5x° ππ + ππ = ππ ππ + ππ − ππ = ππ − ππ ππ = ππ π = ππ B (2x+20)° C ∠π¨πΆπ© = π(ππ)° = ππ° ∠π©πΆπͺ = π(ππ)° + ππ° = ππ° Katrina was solving the problem above and wrote the equation ππ + ππ = ππ. Then, she rewrote this as ππ + ππ = ππ + ππ. Why did she rewrite the equation in this way? How does this help her to find the value of π? b. MP.7 She grouped the quantity on the right-hand side of the equation similarly to that of the left-hand side. This way, it is clear that the quantity ππ on the left-hand side must be equal to the quantity ππ on the right-hand side. Closing (1 minute) ο§ In every unknown angle problem, it is important to identify the angle relationship(s) correctly in order to set up an equation that yields the unknown value. ο§ Check your answer by substituting and/or measuring to be sure it is correct. Lesson Summary ο§ To solve an unknown angle problem, identify the angle relationship(s) first to set up an equation that will yield the unknown value. ο§ Angles on a line and supplementary angles are not the same relationship. Supplementary angles are two angles whose angle measures sum to πππ° whereas angles on a line are two or more adjacent angles whose angle measures sum to πππ°. Exit Ticket (3 minutes) Lesson 2: Solving for Unknown Angles Using Equations This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M6-TE-1.3.0-10.2015 26 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM Name 7•6 Date Lesson 2: Solving for Unknown Angles Using Equations Exit Ticket Two lines meet at a point that is also the vertex of an angle. Set up and solve an equation to find the value of π₯. Explain why your answer is reasonable. 27° 65° x° Lesson 2: Solving for Unknown Angles Using Equations This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M6-TE-1.3.0-10.2015 27 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM 7•6 Exit Ticket Sample Solutions Two lines meet at a point that is also the vertex of an angle. Set up and solve an equation to find the value of π. Explain why your answer is reasonable. ππ + (ππ − ππ) = π π = πππ 65° OR 27° πΛ x° π + ππ = ππ π + ππ − ππ = ππ − ππ π = ππ ππ + π = π ππ + (ππ) = π π = πππ The answers seem reasonable because a rounded value of π as ππ and a rounded value of its adjacent angle ππ as ππ yields a sum of πππ, which is close to the calculated answer. Problem Set Sample Solutions Note: Arcs indicating unknown angles begin to be dropped from the diagrams. It is necessary for students to determine the specific angle whose measure is being sought. Students should draw their own arcs. 1. Two lines meet at a point that is also the endpoint of a ray. Set up and solve an equation to find the value of π. π + ππ + ππ = πππ ∠s on a line π + πππ = πππ Scaffolding: π + πππ − πππ = πππ − πππ c° π = ππ Scaffolded solutions: 17° Students struggling to organize their solution may benefit from prompts such as the following: Write an equation to model this situation. Explain how your equation describes the situation. Solve and interpret the solution. Is it reasonable? a. Use the equation above. b. The angle marked π°, the right angle, and the angle with measurement ππ° are angles on a line, and their measurements sum to πππ°. c. Use the solution above. The answer seems reasonable because it looks like it has a measurement a little less than a ππ° angle. Lesson 2: Solving for Unknown Angles Using Equations This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M6-TE-1.3.0-10.2015 28 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM 2. 7•6 Two lines meet at a point that is also the endpoint of a ray. Set up and solve an equation to find the value of π. Explain why your answer is reasonable. π + ππ = ππ ∠s add and vert. ∠s 49° π + ππ − ππ = ππ − ππ π = ππ The answers seem reasonable because a rounded value of π as ππ and a rounded value of its adjacent angle ππ as ππ yields a sum of ππ, which is close to the rounded value of the measurement of the vertical angle. 33° 3. a° Two lines meet at a point that is also the endpoint of a ray. Set up and solve an equation to find the value of π. π + ππ = πππ ∠s add and vert. ∠s π + ππ − ππ = πππ − ππ π = ππ 4. Two lines meet at a point that is also the vertex of an angle. Set up and solve an equation to find the value of π. (ππ − ππ) + ππ = π ∠s add and vert. ∠s π = ππ m° 24° 68° 5. Three lines meet at a point. Set up and solve an equation to find the value of π. π + πππ + ππ = πππ ∠s on a line and vert. ∠s π + πππ = πππ π + πππ − πππ = πππ − πππ 34° π = ππ r° Lesson 2: Solving for Unknown Angles Using Equations This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M6-TE-1.3.0-10.2015 122° 29 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM 6. 7•6 Three lines meet at a point that is also the endpoint of a ray. Set up and solve an equation to find the value of each variable in the diagram. π = ππ − ππ Complementary ∠s π = ππ π + ππ + ππ + ππ = πππ ∠s on a line 43° 51° π + πππ = πππ π + πππ − πππ = πππ − πππ π = ππ + ππ w° Vert. ∠s π = ππ 7. z° v° π = ππ π = ππ Vert. ∠s π = ππ Vert. ∠s y° x° Set up and solve an equation to find the value of π. Find the measurement of ∠π¨πΆπ© and of ∠π©πΆπͺ. Scaffolding: (ππ − ππ) + πππ = πππ Supplementary ∠s πππ − ππ = πππ πππ − ππ + ππ = πππ + ππ πππ = πππ π = ππ The measurement of ∠π¨πΆπ©: π(ππ)° − ππ° = ππ° The measurement of ∠π©πΆπͺ: ππ(ππ)° = πππ° Students struggling to organize their solution may benefit from prompts such as the following: ο§ Write an equation to model this situation. Explain how your equation describes the situation. Solve and interpret the solution. Is it reasonable? Scaffolded solutions: 8. a. Use the equation above. b. The marked angles are angles on a line, and their measurements sum to πππ°. c. Once ππ is substituted for π, then the measurement of ∠π¨πΆπ© is ππ° and the measurement of ∠π©πΆπͺ is πππ°. These answers seem reasonable since ∠π¨πΆπ© is acute and ∠π©πΆπͺ is obtuse. Set up and solve an equation to find the value of π. Find the measurement of ∠π¨πΆπ© and of ∠π©πΆπͺ. π + π + ππ = ππ Complementary ∠s ππ + π = ππ ππ + π − π = ππ − π ππ = ππ π = ππ π π π π π π The measurement of ∠π¨πΆπ©: (ππ ) ° + π° = ππ ° π π π π The measurement of ∠π©πΆπͺ: π (ππ ) ° = ππ ° Lesson 2: Solving for Unknown Angles Using Equations This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M6-TE-1.3.0-10.2015 30 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM 9. 7•6 Set up and solve an equation to find the value of π. Find the measurement of ∠π¨πΆπ© and of ∠π©πΆπͺ. ππ + π + ππ + ππ = πππ ∠s on a line B ππ + ππ = πππ ππ + ππ − ππ = πππ − ππ ππ = πππ (4x+5)° π = ππ The measurement of ∠π¨πΆπ©: π(ππ)° + π° = ππ° A (5x+22)° O C The measurement of ∠π©πΆπͺ: π(ππ)° + ππ° = πππ° 10. Write a verbal problem that models the following diagram. Then, solve for the two angles. One possible response: Two angles are supplementary. The measurement of one angle is five times the measurement of the other. Find the measurements of both angles. πππ + ππ = πππ Supplementary ∠s 10x° 2x° πππ = πππ π = ππ The measurement of Angle 1: ππ(ππ)° = πππ° The measurement of Angle 2: π(ππ)° = ππ° Lesson 2: Solving for Unknown Angles Using Equations This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G7-M6-TE-1.3.0-10.2015 31 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
