This is an isosceles triangle.
Draw a line from ∠A to the midpoint of segment BC and label that point M. This creates a 90 degree triangle AMC.
∠A for triangle ABC is 20 degrees. Since we bisected ∠A to create segment AM, then ∠A for triangle AMC is 10 degrees.
Since M is the midpoint of segment BC, then segment AC is 1/2 of segment BC (1/2 of 4 cm) = 2 cm.
Now you can use sin x = opposite/hypotenuse
sin ∠A = MC/AC
sin 10 = 2/AC
AC = 2/sin10
AC ≈ -3.68
Note: Since length cannot be negative, this cannot be a true triangle
