The sides of the parallelogram are 3 and 5 cm, the acute angle is 60 degrees. find the area of the parallelogram.

ABCD – parallelogram, AB = CD = 3 cm, AD = BC = 5 cm, angle DAB = 60 degrees.
From the top B we draw the height BH to the side AD. Consider a triangle BHA: angle BHA = 90 degrees, since BH is height, angle HAB (angle DAB) = 60 degrees, AB = 3 cm – hypotenuse, HA and BH are legs. By the theorem on the sum of the angles of a triangle, we find the degree measure of the angle ABH:
angle ABN + angle BHA + angle HAB = 180 degrees;
angle ABH + 90 + 60 = 180;
angle ABH = 180 – 150;
angle ABH = 30 degrees.
In a right-angled triangle, opposite an angle of 30 degrees, lies a leg, which is 2 times less than the hypotenuse, then:
HA = AB / 2 = 3/2 (cm).
By the Pythagorean theorem:
BH = √ (AB ^ 2 – HA ^ 2) = √ (3 ^ 2 – (3/2) ^ 2) = √ (9 – 9/4) = √ ((36 – 9) / 4) = √ ( 27/4) = √27 / √4 = 3√3 / 2 (cm).
The area of ​​the parallelogram is:
S = a * h,
where a is the side of the parallelogram, h is the height drawn to the side a.
S = AD * BH = 5 * 3√3 / 2 = 15√3 / 2 (cm ^ 2).
Answer: S = 15√3 / 2 cm ^ 2.