Find the value of the polygon. the polygon has 6 interior angles 2 of them are y and 4 of them are (2y-20)

First we need to determine what the 6 angles must add to. Turns out we use this formula

S = 180(n-2)

where S is the sum of the angles (result of adding them all up) and n is the number of sides. In this case, n = 6. So let's plug that in to get

S = 180(n-2)
S = 180(6-2)
S = 180(4)
S = 720

The six angles, whatever they are individually, add to 720 degrees. The six angles are y, y, 2y-20, 2y-20,  2y-20,  2y-20,  

They add up and must be equal to 720, so let's set up the equation to get...

(y)+(y)+(2y-20)+(2y-20)+(2y-20)+(2y-20) = 720

Let's solve for y

y+y+2y-20+2y-20+2y-20+2y-20 = 720

10y-80 = 720

10y-80+80 = 720+80

10y = 800

10y/10 = 800/10

y = 80

Now that we know the value of y, we can figure out the six angles

angle1 = y = 80 degrees
angle2 = y = 80 degrees
angle3 = 2y-20 = 2*80-20 = 140 degrees
angle4 = 2y-20 = 2*80-20 = 140 degrees
angle5 = 2y-20 = 2*80-20 = 140 degrees
angle6 = 2y-20 = 2*80-20 = 140 degrees

and that's all there is to it

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)