# The base of the straight prism ABCDA1B1C1D1 is a square ABCD with a side length of 4√5.

**The base of the straight prism ABCDA1B1C1D1 is a square ABCD with a side length of 4√5. The height of the prism is 4√15 Find the cosine of the angle between the plane of the base of the prism and the plane AMD1, where point M is the midpoint of the edge CD.**

Since point M is the middle of CD, then DМ = CD / 2 = 4 * √5 / 2 = 2 * √5 cm.

In a right-angled triangle ADM, according to the Pythagorean theorem, AM ^ 2 = AD ^ 2 + DM ^ 2 = 80 + 20 = 100.

AM = 10 cm.

The area of the triangle ADM is equal to: Sadm = AD * DM / 2 = 4 * √5 * 2 * √5 / 2 = 20 cm2.

Also Sadm = AM * DH / 2.

DН = 2 * Sadm / AM = 2 * 20/10 = 4 cm.

In a right-angled triangle DD1H, D1H2 = DH ^ 2 + DD1 ^ 2 = 16 + 240 = 256.

D1H = 16 cm.

In a right-angled triangle DD1H CosDHD1 = DH / D1H = 4/16 = 1/4 = 0.25.

Answer: The cosine of the angle between the planes is 0.25.