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The measures of angles determined using the angle addition postulate and the angle bisector theorem are:

1. [tex]m \angle JIH = 145^{\circ}[/tex]

2. [tex]\\m \angle IJK = 132^{\circ}[/tex]

3. [tex]m \angle IBA = 120^{\circ}[/tex]

4. [tex]m \angle VUE= 20^{\circ}[/tex]

5. [tex]x = 4[/tex]

6. [tex]x = -6[/tex]

7. [tex]m \angle UVY = 30^{\circ}[/tex]

8. [tex]m \angle IHG = 129^{\circ}[/tex]

1. Find [tex]m \angle JIH[/tex]

  • Given:

[tex]m \angle JIE = 60^{\circ}\\m \angle EIH = 85^{\circ}[/tex]

  • Therefore:

[tex]m \angle JIH = m \angle JIE + m\angle EIH[/tex] (angle addition postulate)

Substitute

[tex]m \angle JIH = 60 + 85\\m \angle JIH = 145^{\circ}[/tex]

2. Find [tex]m \angle IJK[/tex]

  • Given:

[tex]m \angle IJB = 82^{\circ}\\m \angle BJK = 50^{\circ}[/tex]

  • Therefore:

[tex]m \angle IJK = m \angle IJB + m\angle BJK[/tex] (angle addition postulate)

Substitute

[tex]m \angle IJK= 82 + 50\\m \angle IJK = 132^{\circ}[/tex]

3. Find [tex]m \angle IBA[/tex]

  • Given:

[tex]m \angle CBI = 46^{\circ}\\m \angle CBA = 166^{\circ}[/tex]

  • Therefore:

[tex]m \angle CBA = m \angle IBA + m\angle CBI[/tex] (angle addition postulate)

Substitute

[tex]166 = m \angle IBA + 46[/tex]

Subtract 46 from both sides

[tex]166 - 46 = m \angle IBA \\120 = m \angle IBA\\m \angle IBA = 120^{\circ}[/tex]

4. Find  [tex]m \angle VUE[/tex]

  • Given:

[tex]m \angle VUT = 160^{\circ}\\m \angle EUT = 140^{\circ}[/tex]

  • Therefore:

[tex]m \angle VUT= m \angle EUT+ m\angle VUE[/tex] (angle addition postulate)

Substitute

[tex]160 = 140 + m \angle VUE[/tex]

Subtract 140 from both sides

[tex]160 - 140 = m \angle VUE\\20 = m \angle VUE\\m \angle VUE= 20^{\circ}[/tex]

5. Find x

  • Given:

[tex]m \angle PQR = 86^{\circ}\\m \angle BQR = 11x + 6\\m \angle PQB = 10x + 4[/tex]

  • Thus:

[tex]m \angle BQR + m \angle PQB = m \angle PQR[/tex] (Angle addition postulate)

Substitute

[tex]11x + 6 + 10x - 4 = 86\\\[/tex]

Add like terms and solve for x

[tex]21x + 2 = 86\\21x = 86 - 2\\21x = 84\\x = 4[/tex]

6.  Find x

  • Given:

[tex]m \angle LMN= 160^{\circ}\\m \angle RMN= x + 46\\m \angle LMR= x + 126[/tex]

  • Thus:

[tex]m \angle LMR+ m \angle RMN= m \angle LMN[/tex] (Angle addition postulate)

Substitute

[tex]x + 126 + x + 46 = 160[/tex]

Add like terms and solve for x

[tex]2x + 172 = 160[/tex]

[tex]2x = 160 - 172\\2x = - 12\\x = -6[/tex]

7. Find [tex]m \angle UVY[/tex]

  • Given:

[tex]m \angle YVW = 14x + 5\\m \angle UVY = 10 + 2x\\m \angle UVW = 175\\[/tex]

  • First, create an equation to find the value of x.

[tex]14x + 5 + 10 + 2x = 175[/tex] (angle addition postulate)

Add like terms and solve for x

[tex]16x + 15 = 175\\16x = 175 - 15\\16x = 160\\x = 10[/tex]

  • Thus:

[tex]m \angle UVY = 10 + 2x[/tex]

Plug in the value of x

[tex]m \angle UVY = 10 + 2(10) \\m \angle UVY = 30^{\circ}[/tex]

8. Find [tex]m \angle IHG[/tex]

  • Given:

[tex]m \angle IHG = 3 + 21x\\m \angle KHG = 64\\m \angle IHK = 10x + 5[/tex]

  • First, create an equation to find the value of x.

[tex]64 + 10x + 5 = 3 + 21x[/tex] (angle addition postulate)

Add like terms and solve for x

[tex]64 + 10x + 5 = 3 + 21x\\69 + 10x = 3 + 21x\\69 - 3 = -10x +21x\\66 = 11x\\6 = x\\x = 6[/tex]

  • Thus:

[tex]m \angle IHG = 3 + 21x\\[/tex]

Plug in the value of x

[tex]m \angle IHG = 3 + 21(6)\\m \angle IHG = 129^{\circ}[/tex]

The answers are:

1. [tex]m \angle JIH = 145^{\circ}[/tex]

2. [tex]\\m \angle IJK = 132^{\circ}[/tex]

3. [tex]m \angle IBA = 120^{\circ}[/tex]

4. [tex]m \angle VUE= 20^{\circ}[/tex]

5. [tex]x = 4[/tex]

6. [tex]x = -6[/tex]

7. [tex]m \angle UVY = 30^{\circ}[/tex]

8. [tex]m \angle IHG = 129^{\circ}[/tex]

Learn more here:

https://brainly.com/question/24529637

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