Given: ABCD – parallelogram angle ABK = 40 degrees BK – height Find: angles of parallelogram ABCD.

1) Consider a triangle ВAK – (rectangular). In it: AB – hypotenuse, angle ABK = 40gr, angle BKA = 90gr. From here we can find angle A.
angle A = 180 – angle ABK – angle ВKA
angle A = 180 – 90 – 40 = 50g. (by the theorem on the sum of the angles of a triangle)
2) Because AВСD is a parallelogram, then angle C = angle A and is equal to 50gr, angle D = angle B.
3) angle D = 180 – angle C
angle D = 180 -50 = 130gr (according to the property of the sum of the angles of the parallelogram adjacent to one side)
4) Because. angle D = angle B, then angle B = 130gr.