Point O was taken in square ABCD so that triangle BOC is equilateral. Calculate the degree measure of the OAD angle.

Let’s connect point O with the vertices of the square A and D. By the condition of the problem, the triangle BOC is equilateral and:

| BO | = | BC | = | AB |;

∠OBC = 60 °;

Triangles ABO and DCO are isosceles because | BO | = | AB | = | CO | = | СD |:

∠ABO = 90 ° – ∠OBS = 90 ° – 60 ° = 30 °;

∠BAO = ∠BOA = (180 ° – ∠ABO) / 2 = (180 ° – 30 °) / 2 = 75 °;

∠DCO = 90 ° – ∠OCВ = 90 ° – 60 ° = 30 °;

∠COD = ∠CDО = (180 ° – ∠DCO) / 2 = (180 ° – 30 °) / 2 = 75 °;

For ∠OAD we get:

∠OAD = 360 ° – ∠BOC – ∠BOA – ∠COD = 360 ° – 60 ° – 75 ° – 75 ° = 150 °

Answer: 150 °