Two altitudes of an isosceles triangle are equal to 20 cm and 30 cm. determine the possible measures of the base angles of the triangle

Given the image attached, where the altitude 30 is the one in red, 20 the one in green.

CE / AD = BC / AB
20 / 30 = 2 / 3

BC = 2 / 3 AB
BD = 1 / 2 BC = 1 / 3 AB

Using Pythgorean theorem
BD^2 + AD^2 = AB^2
(1/3AB)^2 + 30^2 = AB^2
(1/9AB)^2 + 900 = AB^2
AB^2 + 8100 = 9AB^2
8AB^2 = 8100
AB^2 = (8100/8)
AB = sqrt(8100/8)
AB = 45 / sqrt(2)

Looking for the base angle
sin B = AD/AB
sin B = 30 / [45/sqrt(2)]
sin B = 0.9426090416
B = sin^-1 (0.9426090416)
B = 70.53 degrees

I'm not sure but i think it's 1670km/h