Find the angle at the base of an isosceles triangle whose side length is 2√3 and the area is 3 √3.

Area = 1/2 product of the sides by the sine of the angle between them. Suppose in triangle ABC sides AB and BC are equal, so AB = BC = 2√3. If the area is 3√3, we can find the sine of angle B: sin B = 3√3 / (1/2 * 2√3 * 2√3) = 3√3 / 6 = √3 / 2 => angle B = 60 degrees In an isosceles triangle, the angles at the base are equal, so the angle A = C = (180 – 60) / 2 = 120/2 = 60 degrees
Answer: 60 degrees