Find the angles of the trapezoid in trapezoid ABCD ∠D≈60 °, AB≈BC and AC⊥CD.

By the property of angles in a trapezoid: the sum of the angles belonging to one side is equal to 180.

We have:

∠C + ∠D = 180;

∠С = 180 – 60 = 120.

∠С = ∠ВСА + ∠АСD, where ∠АСD = 90, because AC⊥CD.

120 = ∠BCA + 90; ∠BCA = 30.

ΔABS is an isosceles triangle, because AB = BC.

Then ∠BAC = ∠BCA = 30.

ΔАСD – right-angled triangle and ∠D + ∠АСD + ∠САD = 180,

whence ∠САD = 180 – 60 – 90 = 30.

∠A = ∠BAC + ∠CAD = 30 + 30 = 60.

∠В = 360 – ∠А – ∠С – D = 360 – 60 – 120 – 60 = 120.

Answer: ∠А = ∠D = 60 °, ∠В = ∠С = 120 °.