Find mWYZ as well Find mACB please help with both. Thank you every much! Triangles, Need help, right now I’m does adding more stuff so it can stop saying 20 characters.

m<WYZ = 23°

m<ACB = 87°

Step-by-step explanation:

Problem 1: Find WYZ

Given,

m<WVX = (3x - 7)°

m<VYZ = (16x - 3)°

∆WYZ ~ ∆WVX, therefore:

m<WYZ = m<WVX (corresponding angles of similar triangles are congruent)

m<WYZ = (3x - 7)° (substitution)

Create an equation to find the value of x.

m<WYZ + m<VYZ = 180° (linear pair)

(3x -7)° + (16x - 3)° = 180° (substitution)

Solve for x

3x - 7 + 16x - 3 = 180

Add like terms

19x - 10 = 180

Add 10 to both sides

19x - 10 + 10 = 180 + 10

19x = 190

Divide both sides by 19

19x/19 = 190/19

x = 10

m<WYZ = (3x - 7)

Substitute x = 10

m<WYZ = 3(10) - 7 = 30 - 7

m<WYZ = 23°

Problem 2: Find m<ACB

Given,

m<A = 62°

m<AED = (11x - 2)°

m<B = (6x + 13)°

∆ADE ~ ∆ACB, therefore:

m<AED = m<B (corresponding angles of similar ∆s are congruent)

(11x - 2)° = (6x + 13)°

Solve for x

11x - 2 = 6x + 13

Collect like terms

11x - 6x = 2 + 13

5x = 15

Divide both sides by 5

5x/5 = 15/5

x = 3

m<ACB + m<B + m<A = 180° (sum of ∆)

m<ACB + (6x + 13) + 62° = 180° (substitution)

Plug in the value of x and solve

m<ACB + (6(3) + 13) + 62° = 180°

m<ACB + (6(3) + 13) + 62° = 180°

m<ACB + (18 + 13) + 62° = 180°

m<ACB + 31° + 62° = 180°

m<ACB + 93° = 180°

Subtract 93 from both sides

m<ACB = 180° - 93°

m<ACB = 87°