A plane is drawn through the leg of an isosceles right-angled triangle, which forms an angle of 60 with the plane

A plane is drawn through the leg of an isosceles right-angled triangle, which forms an angle of 60 with the plane of the triangle. Find the angles that form the other two sides of the triangle with this plane.

Rectangle ABC is rectangular with a right angle at the vertex C. Segment AC is a leg of the triangle and lies on the plane α, since AB is perpendicular to AC, then the projection CH is perpendicular to the segment AC, and therefore the angle BCH is the angle between the plane of the triangle ABC and the plane α which is 60.

The ABC triangle is isosceles and rectangular, then the angle BAC = ABC = 45.

Let the legs AC and BC be equal to X.

Then the hypotenuse X / AB = sin 45 °.

AB = X / sin 45 ° = X / 1 / √2 = X * √2.

In a triangle СBН, BН / СB = sin 60 °.

BH = X * (√3 / 2).

Then SinBAH = BH / AB = (X * (√3 / 2)) / (X * √2) = √3 / (2 * √2).

Angle BAN = arcsin √3 / (2 * √2) = arcsin √6 / 4.

Answer: Angle BCH = 60, angle BAN = arcsin √6 / 4.