The bases of an isosceles trapezoid are 6 and 12. The sine of an acute angle of a trapezoid is 0.8. Find the side.

Let us introduce the notation, let ABCD be a given trapezoid (AB = CD), BC = 6, AD = 12, sinA = 0.8.

Let us omit the heights BH and CE from the obtuse corners of the trapezoid. The BHC quadrilateral is a rectangle, so НЕ = BC = 6.

Triangles ABH and DCE are equal (right-angled triangles with equal hypotenuses and equal angles A and D), so AH = ED = (12 – 6): 2 = 3.

Let’s calculate the cosine of the angle A: since sin²x + cos²x = 1, then cosx = √ (1 – sin²x).

cosA = √ (1 – 0.64) = √0.36 = 0.6.

In triangle ABH: cosA = AH / AB; 0.6 = 3 / AB; AB = 3 / 0.6 = 5.

Answer: the side of the trapezoid is 5.