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tramserran
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

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