Find the diagonal AC of the parallelogram ABCD, if AB is 3 cm, BC is 4 cm, angle B is 150 degrees.

Consider a triangle ABC: AB = 3 cm, BC = 4 cm, ∠ B = 150 °.

The side AC, opposite the angle B, can be found by the cosine theorem: the square of the side of a triangle is equal to the sum of the squares of the other two sides minus their double product by the cosine of the angle between them.

AC ^ 2 = AB ^ 2 + BC ^ 2 – 2 * AB * BC * cos B = 32 + 42 – 2 * 3 * 4 * cos 150 °;

AC ^ 2 = 9 + 16 – 24 * (- √3 / 2) = 25 + 12√3 ≈ 45.78;

AC = √45.78 ≈ 6.77 cm.

The AC diagonal of the parallelogram ABCD is 6.77 cm.