Question

13 POINTS- please help me

Answer 1

See explanation

Explanation:

16. Two parallel lines are cut by transversal. Angles with measures
`(6x+20)^(\circ)` and
`(x+100)^(\circ)` are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

`6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16`

Then

`(6x+20)^(\circ)=(6\cdot 16+20)^(\circ)=116^(\circ)\\ \\(x+100)^(\circ)=(16+100)^(\circ)=116^(\circ)`

17. Two parallel lines are cut by transversal. Angles with measures
`(2x+12)^(\circ)` and
`(3x-22)^(\circ)` are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

`2x+12=3x-22\\ \\2x-3x=-22-12\\ \\-x=-34\\ \\x=34`

Then

`(2x+12)^(\circ)=(2\cdot 34+12)^(\circ)=80^(\circ)\\ \\(3x-22)^(\circ)=(3\cdot 34-22)^(\circ)=80^(\circ)`

18. Two parallel lines are cut by transversal. Angles with measures
`(6x-7)^(\circ)` and
`(5x+10)^(\circ)` are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

`6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17`

Then

`(6x-7)^(\circ)=(6\cdot 17-7)^(\circ)=95^(\circ)\\ \\(5x+10)^(\circ)=(5\cdot 17+10)^(\circ)=95^(\circ)`

19. The diagram shows two complementary angles with measures
`2x^(\circ)` and
`56^(\circ)`. The measures of complementary angles add up to
`90^(\circ),` then

`2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17`

Hence,

`2x^(\circ)=2\cdot 17^(\circ)=34^(\circ)`

Check:

`34^(\circ)+56^(\circ)=90^(\circ)`

20. Angles
`\angle 1` and
`\angle 2` are vertical angles. By vertical angles theorem, vertical angles are congruent, so

`m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4`

Hence,

`m\angle 1=(5x+7)^(\circ)=(5\cdot 4+7)^(\circ)=27^(\circ)\\ \\m\angle 2=(3x+15)^(\circ)=(3\cdot 4+15)^(\circ)=27^(\circ)`

21.
`\angle 5` and
`\angle 8` are supplementary. The measures of supplementary angles add up to
`180^(\circ),` so

`m\angle 5+m\angle 8=180^(\circ)\\ \\3x-40+7x-120=180\\ \\10x-160=180\\ \\10x=180+160\\ \\10x=340\\ \\x=34`

Therefore,

`m\angle 5=(3x-40)^(\circ)=(3\cdot 34-40)^(\circ)=62^(\circ)\\ \\m\angle 8=(7x-120)^(\circ)=(7\cdot 34-120)^(\circ)=118^(\circ)\\ \\62^(\circ)+118^(\circ)=180^(\circ)`