Question

b0rf

Answer 1

@563blackghost

Answer 2

@563blackghost @3mar

Answer 3

Are you persuaded of the last one?

Answer 4

a bit yeah

Answer 5

Use the question we helped with last time. If you look at the two we proved `Angle 3 and 6` are congruent based off the `Alternate Exterior Angles Theorem` which closely relates to your problem. So...

\(\huge\bf{2x+50=5x+20}\)

Answer 6

x=10

Answer 7

Nice :)

Answer 8

my mother is calling me!
I will be back within 15 min In Sha' Allah
SALAM!

Answer 9

The two angles lie on the same line but are on opposite sides of parallel transversals. Due to this they would have a sum of 180.

\(\huge\bf{(4x+20)+(x+10)=180}\)

Answer 10

i would guess x is 35?

Answer 11

Not quite. Combine like terms.

\(\huge\bf{5x+30=180}\)

Subtract 30 from each side.

\(\huge\bf{5x=150}\)

Divide by 5.

\(\huge\bf{\frac{5x}{5}=\frac{150}{5}}\)

Answer 12

30

Answer 13

Nice :)

Answer 14

"The two angles lie on the same line but are on opposite sides of parallel transversals"
This is what is called: \(\LARGE \color{Crimson }{Consecutive ~~Exterior ~~Angles.}\)

We were searching for this types of angles at the previous problem!
- angles 3 and 8
- angles 1 and 6