Can anyone please do these questions? This is Unit 1 Geometry basics Homework 5 Angle Relationships.

Answer:

Step-by-step explanation:

11). 6x + 7 = 8x - 17 [Since vertical angles are equal in measure]

    8x - 6x = 17 + 7

    2x = 24

    x = 12

12). (11x - 15) + (5x - 13) = 180°

    16x - 28 = 180°

    16x = 180 + 28

    16x = 208

    x = 13

13). Since DB⊥AC,

    m∠CBD = 90°

    m∠CBE + m∠DBE = 90°

    (5x - 42) + (2x - 1) = 90°

     7x - 43 = 90

    7x = 133

     x = 19

14). Since QS bisects angle PQR,

    m∠PQS = m∠RQS

    10x + 1 = [tex]\frac{82}{2}[/tex]

    10x + 1 = 41

    10x = 40

    x = 4

15). 10x - 61 = 18y + 5 [Vertical angles]

     10x - 18y = 61 + 5

     10x - 18y = 66

     5x - 9y = 33 ------(1)

     (18y + 5) + (x + 10) = 180 [linear pair of angles are supplementary]

     18y + x + 15 = 180

     x + 18y = 165 ------(2)

     By adding equation (1)×2 and equation (2)

     (10x - 18y) + (x + 18y) = 66 + 165

     11x = 231

      x = 21

16). (5x - 17) + (3x - 11) = 180 [[linear pair of angles are supplementary]

      8x - 28 = 180

      8x = 180 + 28

      8x = 208

      x = 26

      (3x - 11)° = 78 - 11

                   = 67°

      67° + 90° + (2y + 5)°= 180° [Sum of angles on a line]

      162 + 2y = 180

      2y = 180 - 162

      2y = 18

      y = 9

17). NP bisects ∠MNQ.

    Therefore, m∠MNQ = 2(m∠PNQ)

    8x + 12 = 2×78

    8x + 12 = 156

    8x = 156 - 12

    8x = 144

    x = 18

    m∠MNQ = (8x + 12)° = 156°

    m∠RNM = m∠ONQ [Vertical angles]

    (3y - 9)° = 180° - m∠MNQ

    3y - 9 = 180 - 156

    3y - 9 = 24

    3y = 33

    y = 11

Using the knowledge of the angle relationships to create an equation, the value of x and y in the image given are:

11. x = 12

12. x = 13

13. x = 19

14. x = 4

15. x = 21; y = 8

16. x = 26; y = 9

17. x = 18; y = 11

Applying the knowledge of angle relationships, we can solve each given problem by first, creating a an equation from the relationship between angles, then solve for the variables.

11. [tex]6x + 7 $ and $ 8x - 17[/tex] are vertical angles.

[tex]6x + 7 = 8x - 17[/tex]

[tex]6x - 8x = -7 - 17\\\\-2x = -24\\\\x = 12[/tex]

12. [tex]11x - 15 $ and $ 5x - 13[/tex] are linear pair angles.

[tex](11x - 15) + (5x - 13) = 180[/tex]

[tex]11x - 15 + 5x - 13 = 180\\\\16x - 28 = 180\\\\16x - 28 + 28 = 180 + 28\\\\16x = 208\\\\x = 13[/tex]

13. <DBE and <CBE are complementary angles.

[tex]m \angle DBE + m \angle CBE = 90[/tex]

[tex](2x - 1) + (5x - 42) = 90[/tex]

[tex]2x - 1 + 5x - 42 = 90\\\\7x - 43 = 90\\\\7x = 90 + 43\\\\7x = 133\\\\x = 19[/tex]

14. Since QS bisects <PQR, therefore,

m<PQS = 1/2(PQR)

[tex]10x + 1 = \frac{1}{2} (82)[/tex]

[tex]10x + 1 = 41\\\\10x = 41 - 1\\\\10x = 40\\\\x = 4[/tex]

15. [tex](x + 10) $ and $ (10x - 16)[/tex] are linear pair angles.

[tex](x + 10) + (10x - 16) = 180\\[/tex]

[tex]x + 10 + 10x - 61 = 180\\\\11x - 51 = 180\\\\11x = 180 + 51\\\\11x = 231\\\\x = 21[/tex]

Since [tex](10x - 61) $ and $ (10y + 5)[/tex] are vertical angles, they are congruent. Therefore,

[tex](10x - 61) = (10y + 5)[/tex]

[tex]10(21) - 61 = 18y + 5\\\\149 = 18y + 5\\\\149 - 5 = 10y\\\\144 = 18y\\\\8 = y\\\\\mathbf{y = 8}[/tex]

16. Since [tex](5x - 17) $ and $ (3x - 11)[/tex] are linear pair, their sum equals 180 degrees. Therefore,

[tex](5x - 17) + (3x - 11) = 180[/tex]

[tex]5x - 17 + 3x - 11 = 180\\\\8x - 28 = 180\\\\8x = 180 + 28\\\\8x = 208\\\\x = 26[/tex]

[tex](2y + 5) + 90 + (3x - 11) = 180[/tex] (angles on a straight line = 180 degrees)

[tex]2y + 5 + 90 + 3(26) - 11 = 180\\\\2y + 5 + 90 + 78 - 11 = 180\\\\2y + 162 = 180\\\\2y = 180 - 162\\\\2y =18\\\\y = 9[/tex]

17. Since NP bisects <MNQ, therefore,

[tex]m \angle PNQ = \frac{1}{2}(m \angle MNQ)[/tex]

[tex]78 = \frac{1}{2}(8x + 12)[/tex]

[tex]78 \times 2 = 8x + 12\\\\156 = 8x + 12\\\\156 - 12 = 8x \\\\144 = 8x\\\\18 = x\\\\x = 18[/tex]

[tex]m \angle RNM + m \angle MNQ = 180^{\circ}[/tex] (linear pair)

[tex](3y - 9) + (8x + 12) = 180[/tex]

[tex]3y - 9 + 8(18) + 12 = 180\\\\3y - 9 + 144 + 12 = 180\\\\3y + 147 = 180\\\\3y = 180 - 147\\\\3y = 33\\\\y = 11[/tex]

Therefore, using the knowledge of the angle relationships to create an equation, the value of x and y in the image given are:

11. x = 12

12. x = 13

13. x = 19

14. x = 4

15. x = 21; y = 8

16. x = 26; y = 9

17. x = 18; y = 11

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