In triangle ABC, AC = CB = 8, angle ACB = 120 degrees. Point M is removed from the plane of the triangle

In triangle ABC, AC = CB = 8, angle ACB = 120 degrees. Point M is removed from the plane of the triangle at a distance of 12 cm. Point M is removed from the plane of the triangle at a distance of 12 cm and is at an equal distance from the vertices of the triangle ABC. Find the angle between MA and the plane of triangle ABC.

Since point M is at an equal distance from the vertices of the triangle, then its projection onto the plane of the triangle will coincide with the center of the circle circumscribed around this triangle.

Since triangle ABC is isosceles, its angles at the base of AB are equal.

Then the angle BAC = ABC = (180 – 120) / 2 = 300.

Determine the radius of the circumscribed circle.

R = ОА = АС / (2 * SinАВС) = 8/2 * Sin30 = 8 cm.

In a right-angled triangle AOM, the angle OAM is our desired angle, then tgOAM = MO / AO = 12/8 = 3/2.

OAM angle = arctg3 / 2.

Answer: The angle between MA and the plane of the triangle is arctg3 / 2.