ABC triangle, angle C = 90 degrees, angle B = 60 degrees, AM perpendicular (ABC), AM = h

ABC triangle, angle C = 90 degrees, angle B = 60 degrees, AM perpendicular (ABC), AM = h, dihedral angle ABCM = 30 degrees. find the area of triangle MBC.

Since the angle ACB = 90, then the MC is perpendicular to the BC, AC is the projection of the MC onto the plane ABC, then the linear angle of the ACM is equal to the dihedral angle ABCM.
In a right-angled AFM triangle, the AM leg lies opposite an angle of 300, then CM = 2 * AM = 2 * h cm.
By the Pythagorean theorem, AC2 = CM ^ 2 – AM ^ 2 = 4 * h ^ 2 – h ^ 2 = 3 * h ^ 2.
AC = h * √3 cm.
In a right-angled triangle ABC tg60 = AC / BC.
BC = AC / tg60 = h * √3 / √3 = h see.
Then Smvs = MS * BC / 2 = 2 * h * h / 2 = h ^ 2 cm2.
Answer: The area of the MBC triangle is h ^ 2 cm2.