You are given a rectangular parallelepiped ABCDA1B1C1D1. Find the dihedral angle B1ADB

You are given a rectangular parallelepiped ABCDA1B1C1D1. Find the dihedral angle B1ADB if AC = 6√2 m, AB1 = 4√3 m, ABCD-square.

The degree measure of the dihedral angle B1ADB is the linear angle B1AB. He needs to be found. From the bottom: t. ABCD is a square, then AB = BC = x By the Pythagorean theorem: x ^ 2 + x ^ 2 = (6√2) ^ 2 2x ^ 2 = 36 * 2 x ^ 2 = 36 x = ± 6 (-6 extraneous root ). Then the side of the base is 6. Consider a triangle ABB1: It is rectangular, so cosA = 6 / (4√3) = 3 / (2√3) = √3 / 2 => angle A = 30 °