Example 1.
\nOne angle is five degrees less than three times the measure of another angle. Together, the angle measures have a sum of 143\u00b0. What is the measure of each angle?<\/p>\n
Provide students with time to make sense of the problem and persevere in solving it. They could begin their work by guessing and checking, drawing a diagram, or other methods as appropriate. Then, move to the algebraic method shown below.
\n\u2192 What do we need to do first to solve this problem?
\n\u2192 First, we need to define our variable (symbol). Let x be the measure of the first angle in degrees.
\n\u2192 If x is the measure of the first angle, how do you represent the measure of the second angle?
\n\u2192 The second angle is 3x-5.
\n\u2192 What is the equation that represents this situation?
\n\u2192 The equation is x+3x-5=143.
\n\u2192 The equation that represents this situation is x+3x-5=143. Solve for x, and then determine the measure of each angle.<\/p>\n
As students share their solutions for this and subsequent problems, ask them a variety of questions to reinforce the concepts of the last few lessons. For example, ask students to discuss whether or not this is a linear equation and how they know, to justify their steps and explain why they chose their particular first step, to explain what the solution means, or to justify how they know their answer is correct.
\nAnswer:
\nx+3x-5=143
\n(1+3)x-5=143
\n4x-5=143
\n4x-5+5=143+5
\n4x=148
\nx=37
\nThe measure of the first angle is 37\u00b0. The second angle is 3(37\u00b0)-5\u00b0=111\u00b0-5\u00b0=106\u00b0.<\/p>\n
Example 2.
\nGiven a right triangle, find the degree measure of the angles if one angle is ten degrees more than four times the degree measure of the other angle and the third angle is the right angle.<\/p>\n
Give students time to work. As they work, walk around and identify students who are writing and solving the problem in different ways. The instructional goal of this example is to make clear that there are different ways to solve a linear equation as opposed to one \u201cright way.\u201d Select students to share their work with the class. If students do not come up with different ways of solving the equation, talk them through the following student work samples.
\nAgain, as students share their solutions, ask them a variety of questions to reinforce the concepts of the last few lessons. For example, ask students to discuss whether or not this is a linear equation and how they know, to justify their steps and explain why they chose their particular first step, to explain what the solution means, or to justify how they know their answer is correct.
\nAnswer:
\nSolution One<\/u>
\nLet x be the measure of the first angle. Then, the second angle is 4x+10. The sum of the measures for the angles for this right triangle is x+4x+10+90=180.
\nx+4x+10+90=180
\n(1+4)x+100=180
\n5x+100=180
\n5x+100-100=180-100
\n5x=80
\nx=16
\nThe measure of the first angle is 16\u00b0, the measure of the second angle is 4(16\u00b0)+10\u00b0=64\u00b0+10\u00b0=74\u00b0, and the measure of the third angle is 90\u00b0.<\/p>\n
Solution Two<\/u>
\nLet x be the measure of the first angle. Then, the second angle is 4x+10. Since we have a right triangle, we already know that one angle is 90\u00b0, which means that the sum of the other two angles is 90: x+4x+10=90.
\nx+4x+10=90
\n(1+4)x+10=90
\n5x+10=90
\n5x+10-10=90-10
\n5x=80
\nx=16
\nThe measure of the first angle is 16\u00b0, the measure of the second angle is 4(16\u00b0)+10\u00b0=64\u00b0+10\u00b0=74\u00b0, and the measure of the third angle is 90\u00b0.<\/p>\n
Solution Three<\/u>
\nLet x be the measure of the second angle. Then, the first angle is \\(\\frac{x-10}{4}\\). Since we have a right triangle, we already know that one angle is 90\u00b0, which means that the sum of the other two angles is 90: x+\\(\\frac{x-10}{4}\\)=90.
\nx+\\(\\frac{x-10}{4}\\)=90
\n4(x+\\(\\frac{x-10}{4}\\)=90)
\n4x+x-10=360
\n(4+1)x-10=360
\n5x-10=360
\n5x-10+10=360+10
\n5x=370
\nx=74
\nThe measure of the second angle is 74\u00b0, the measure of the first angle is \\(\\frac{74^{\\circ}-10^{\\circ}}{4}\\)=\\(\\frac{64^{\\circ}}{4}\\)=16\u00b0, and the measure of the third angle is 90\u00b0.<\/p>\n
Solution Four<\/u>
\nLet x be the measure of the second angle. Then, the first angle is \\(\\frac{x-10}{4}\\). The sum of the three angles is
\nx+\\(\\frac{x-10}{4}\\)+90=180.
\nx+\\(\\frac{x-10}{4}\\)+90=180
\nx+\\(\\frac{x-10}{4}\\)+90-90=180-90
\nx+\\(\\frac{x-10}{4}\\)=90
\n\\(\\frac{4 x}{4}\\)+\\(\\frac{x-10}{4}\\)=90
\n\\(\\frac{4 x+x-10}{4}\\)=90
\n4x+x-10=360
\n5x-10+10=360+10
\n5x=370
\nx=74
\nThe measure of the second angle is 74\u00b0, the measure of the first angle is \\(\\frac{74^{\\circ}-10^{\\circ}}{4}\\)=\\(\\frac{64^{\\circ}}{4}\\)=16\u00b0, and the measure of the third angle is 90\u00b0.<\/p>\n
Make sure students see at least four different methods of solving the problem. Conclude this example with the statements below.
\n\u2192 Each method is slightly different either in terms of how the variable is defined or how the properties of equality are used to solve the equation. The way you find the answer may be different from your classmates\u2019 or your teacher\u2019s.
\n\u2192 As long as you are accurate and do what is mathematically correct, you will find the correct answer.<\/p>\n
Exercises
\nFor each of the following problems, write an equation and solve.<\/p>\n
Exercise 1.
\nA pair of congruent angles are described as follows: The degree measure of one angle is three more than twice a number, and the other angle\u2019s degree measure is 54.5 less than three times the number. Determine the measure of the angles in degrees.
\nAnswer;
\nLet x be the number. Then, the measure of one angle is 3+2x, and the measure of the other angle is 3x-54.5. Because the angles are congruent, their measures are equal. Therefore,
\n3+2x=3x-54.5
\n3+2x-2x=3x-2x-54.5
\n3=x-54.5
\n3+54.5=x-54.5+54.5
\n57.5=x
\nReplacing x with 57.5 in 3+2x gives 3+2(57.5)=3+115=118; therefore the measure of the angles is 118\u00b0.<\/p>\n
Exercise 2.
\nThe measure of one angle is described as twelve more than four times a number. Its supplement is twice as large. Find the measure of each angle in degrees.
\nAnswer;
\nLet x be the number. Then, the measure of one angle is 4x+12. The other angle is 2(4x+12)=8x+24. Since the angles are supplementary, their sum must be 180\u00b0.
\n4x+12+8x+24=180
\n12x+36=180
\n12x+36-36=180-36
\n12x=144
\nx=12
\nReplacing x with 12 in 4x+12 gives 4(12)+12=48+12=60. Replacing x with 12 in 2(4x+12) gives 2(4(12)+12)=2(48+12)=2(60)=120. Therefore, the measures of the angles are 60\u00b0 and 120\u00b0.<\/p>\n
Exercise 3.
\nA triangle has angles described as follows: The measure of the first angle is four more than seven times a number, the measure of the second angle is four less than the first, and the measure of the third angle is twice as large as the first. What is the measure of each angle in degrees?
\nAnswer:
\nLet x be the number. The measure of the first angle is 7x+4. The measure of the second angle is 7x+4-4=7x. The measure of the third angle is 2(7x+4)=14x+8. The sum of the angles of a triangle must be 180\u00b0.
\n7x+4+7x+14x+8=180
\n28x+12=180
\n28x+12-12=180-12
\n28x=168
\nx=6
\nReplacing x with 6 in 7x+4 gives 7(6)+4=42+4=46. Replacing x with 6 in 7x gives 7(6)=42. Replacing x with 6 in 14x+8 gives 14(6)+8=84+8=92. Therefore, the measures of the angles are 46\u00b0, 42\u00b0, and 92\u00b0.<\/p>\n
Exercise 4.
\nOne angle measures nine more than six times a number. A sequence of rigid motions maps the angle onto another angle that is described as being thirty less than nine times the number. What is the measure of the angle in degrees?
\nAnswer:
\nLet x be the number. Then, the measure of one angle is 6x+9. The measure of the other angle is 9x-30. Since rigid motions preserve the measures of angles, then the measure of these angles is equal.
\n6x+9=9x-30
\n6x+9-9=9x-30-9
\n6x=9x-39
\n6x-9x=9x-9x-39
\n-3x=-39
\nx=13
\nReplacing x with 13 in 6x+9 gives 6(13)+9=78+9=87. Therefore, the angle measure is 87\u00b0.<\/p>\n
Exercise 5.
\nA right triangle is described as having an angle of measure six less than negative two times a number, another angle measure that is three less than negative one-fourth the number, and a right angle. What are the measures of the angles in degrees?
\nAnswer;
\nLet x be a number. Then, the measure of one angle is -2x-6. The measure of the other angle is –\\(\\frac{x}{4}\\)-3. The sum of the two angles must be 90\u00b0.
\n-2x-6+(-\\(\\frac{x}{4}\\))-3=90
\n(-\\(\\frac{8x}{4}\\))+(-\\(\\frac{x}{4}\\))-9=90
\n(-\\(\\frac{9x}{4}\\))-9+9=90+9
\n–\\(\\frac{9x}{4}\\)=99
\n-9x=396
\nx=-44
\nReplacing x with -44 gives -2x-6 gives -2(-44)-6=88-6=82. Replacing x with -44 in –\\(\\frac{x}{4}\\)-3 gives 90-82=8. Therefore, the angle measures are 82\u00b0 and 8\u00b0.<\/p>\n
Exercise 6.
\nOne angle is one less than six times the measure of another. The two angles are complementary angles. Find the measure of each angle in degrees.
\nAnswer:
\nLet x be the measure of the first angle. Then, the measure of the second angle is 6x-1. The sum of the measures will be 90 because the angles are complementary.
\nx+6x-1=90
\n7x-1=90
\n7x-1+1=90+1
\n7x=91
\nx=13
\nThe first angle is x and therefore measures 13\u00b0. Replacing x with 13 in 6x-1 gives 6(13)-1=78-1=77. Therefore, the second angle measure is 77\u00b0.<\/p>\n
Students practice writing and solving linear equations.<\/p>\n
For each of the following problems, write an equation and solve.<\/p>\n
Question 1.
\nThe measure of one angle is thirteen less than five times the measure of another angle. The sum of the measures of the two angles is 140\u00b0. Determine the measure of each angle in degrees.
\nAnswer:
\nLet x be the measure of the one angle. Then, the measure of the other angle is 5x-13.
\nx+5x-13=140
\n6x-13=140
\n6x-13+13=140+13
\n6x=153
\nx=25.5
\nSince one angle measure is x, it is 25.5\u00b0. Replacing x with 25.5 in 5x-13 gives
\n5(25.5)-13=140-25.5=114.5. Therefore, the other angle measures 114.5\u00b0.<\/p>\n
Question 2.
\nAn angle measures seventeen more than three times a number. Its supplement is three more than seven times the number. What is the measure of each angle in degrees?
\nAnswer:
\nLet x be the number. Then, the measure of one angle is 3x+17. The measure of the other angle is 7x+3. Since the angles are supplementary, the sum of their measures will be 180.
\n3x+17+7x+3=180
\n10x+20=180
\n10x+20-20=180-20
\n10x=160
\nx=16
\nReplacing x with 16 in 3x+17 gives 3(16)+17=65. Replacing x with 16 in 7x+3 gives (16)+3= 112+3=115. Therefore, the angle measures are 65\u00b0 and 115\u00b0.<\/p>\n
Question 3.
\nThe angles of a triangle are described as follows: \u2220A is the largest angle; its measure is twice the measure of \u2220B. The measure of \u2220C is 2 less than half the measure of \u2220B. Find the measures of the three angles in degrees.
\nAnswer:
\nLet x be the measure of \u2220B. Then, the measure of \u2220A is 2x, and the measure of \u2220C is \\(\\frac{x}{2}\\)-2. The sum of the measures of the angles must be 180\u00b0.
\nx+2x+\\(\\frac{x}{2}\\) -2=180
\n3x+\\(\\frac{x}{2}\\) -2+2=180+2
\n3x+\\(\\frac{x}{2}\\) =182
\n6\\(\\frac{x}{2}\\) +\\(\\frac{x}{2}\\) =182
\n7\\(\\frac{x}{2}\\) =182
\n7x=364
\nx=52
\nSince x is the measure of \u2220B, then \u2220B is 52\u00b0. Replacing x with 52 in 2x gives 2(52)=104. Therefore, the measure of \u2220A is 104\u00b0. Replacing x with 52 in \\(\\frac{x}{2}\\) -2 gives \\(\\frac{52}{2}\\)-2=26-2=24. Therefore, the measure of \u2220C is 24\u00b0.<\/p>\n
Question 4.
\nA pair of corresponding angles are described as follows: The measure of one angle is five less than seven times a number, and the measure of the other angle is eight more than seven times the number. Are the angles congruent? Why or why not?
\nAnswer:
\nLet x be the number. Then, the measure of one angle is 7x-5, and the measure of the other angle is 7x+8. Assume they are congruent, which means their measures are equal.
\n7x-5=7x+8
\n7x-7x-5=7x-7x+8
\n-5\u22608
\nSince -5\u22608, the angles are not congruent.<\/p>\n
Question 5.
\nThe measure of one angle is eleven more than four times a number. Another angle is twice the first angle\u2019s measure. The sum of the measures of the angles is 195\u00b0. What is the measure of each angle in degrees?
\nAnswer:
\nLet x be the number. The measure of one angle can be represented with 4x+11, and the other angle\u2019s measure can be represented as 2(4x+11)=8x+22.
\n4x+11+8x+22=195
\n12x+33=195
\n12x+33-33=195-33
\n12x=162
\nx=13.5
\nReplacing x with 13.5 in 4x+11 gives 4(13.5)+11=54+11=65. Replacing x with 13.5 in 2(4x+11) gives 2(4(13.5)+11)=2(54+11)=2(65)=130. Therefore, the measures of the angles are 65\u00b0 and 130\u00b0.<\/p>\n
Question 6.
\nThree angles are described as follows: \u2220B is half the size of \u2220A. The measure of \u2220C is equal to one less than two times the measure of \u2220B. The sum of \u2220A and \u2220B is 114\u00b0. Can the three angles form a triangle? Why or why not?
\nAnswer:
\nLet x represent the measure of \u2220A. Then, the measure of \u2220B is \\(\\frac{x}{2}\\) , and the measure of \u2220C is 2(\\(\\frac{x}{2}\\) )-1=x-1.
\nThe sum of the measures of \u2220A and \u2220B is 114.
\nx+\\(\\frac{x}{2}\\) =114
\n\\(\\frac{3x}{2}\\) =114
\n3x=228
\nx=76
\nSince x is the measure of \u2220A, then \u2220A is 76\u00b0. Replacing x with 76 in \\(\\frac{x}{2}\\) gives \\(\\frac{76}{2}\\)=38; therefore, the measure of \u2220B is 38\u00b0. Replacing x with 76 in x-1 gives 76-1=75, therefore the measure of \u2220C is 75\u00b0. The sum of the three angle measures is 76\u00b0+38\u00b0+75\u00b0=189\u00b0. Since the sum of the measures of the interior angles of a triangle must equal 180\u00b0, these angles do not form a triangle. The sum is too large.<\/p>\n
For each of the following problems, write an equation and solve.<\/p>\n
Question 1.
\nGiven a right triangle, find the measures of all of the angles, in degrees, if one angle is a right angle and the measure of the second angle is six less than seven times the measure of the third angle.
\nAnswer:
\nLet x represent the measure of the third angle. Then, 7x-6 can represent the measure of the second angle. The sum of the two angles in the right triangle will be 90\u00b0.
\n7x-6+x=90
\n8x-6=90
\n8x-6+6=90+6
\n8x=96
\n\\(\\frac{8}{8}\\) x=\\(\\frac{96}{8}\\)
\nx=12
\nThe third angle is x and therefore measures 12\u00b0. Replacing x with 12 in 7x-6 gives 7(12)-6=84-6=78. Therefore, the measure of the second angle is 78\u00b0. The measure of the third angle is 90\u00b0.<\/p>\n
Question 2.
\nIn a triangle, the measure of the first angle is six times a number. The measure of the second angle is nine less than the first angle. The measure of the third angle is three times the number more than the measure of the first angle. Determine the measure of each angle in degrees.
\nAnswer:
\nLet x be the number. Then, the measure of the first angle is 6x, the measure of the second angle is 6x-9, and the measure of the third angle is 3x+6x. The sum of the measures of the angles in a triangle is 180\u00b0.
\n6x+6x-9+3x+6x=180
\n21x-9=180
\n21x-9+9=180+9
\n21x=189
\n\\(\\frac{21}{21}\\) x=\\(\\frac{189}{21}\\)
\nx=9
\nReplacing x with 9 in 6x gives 6(9)=54. Replacing x with 9 in 6x-9 gives 6(9)-9=54-9=45. Replacing x with 9 in 3x+6x gives 54+3(9)=54+27=81. Therefore, the angle measures are 54\u00b0, 45\u00b0, and 81\u00b0.<\/p>\n","protected":false},"excerpt":{"rendered":"
Engage NY Eureka Math 8th Grade Module 4 Lesson 5 Answer Key Eureka Math Grade 8 Module 4 Lesson 5 Example Answer Key Example 1. One angle is five degrees less than three times the measure of another angle. Together, the angle measures have a sum of 143\u00b0. What is the measure of each angle? … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/85025"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=85025"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/85025\/revisions"}],"predecessor-version":[{"id":137232,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/85025\/revisions\/137232"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=85025"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=85025"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=85025"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}