Consider the figure below. Line FG is parallel to line JK. Part A:Solve for a justify your answer
Part B:Determine m< FNT
Part C:Determine m

A. a = 10

B. m<FNT = 120°

C. m<KTU = 60°

Step-by-step explanation:

A. (7a + 50)° and (14a - 20)° are corresponding angles. Therefore:

(7a + 50)° = (14a - 20)°

Use this equation to find the value of a

7a + 50 = 14a - 20

Combine like terms

7a - 14a = - 50 - 20

-7a = -70

Divide both sides by -7

-7a/-7 = -70/-7

a = 10

B. m<FNT = (14a - 20)° (alternate interior angles are congruent)

Plug in the value of a

m<FNT = 14(10) - 20

m<FNT = 140 - 20

m<FNT = 120°

C. m<KTU + (14a - 20)° = 180° (linear pair)

Plug in the value of a

m<KTU + 14(10) - 20 = 180

m<KTU + 120 = 180

m<KTU + 120 - 120 = 180 - 120

m<KTU = 60°

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

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

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

in δacy and δaby,

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)