Find the dihedral angle between the adjacent side faces of a regular quadrangular pyramid with all edges equal to 1.

Let EABCD be a given quadrangular pyramid (E is a top, ABCD is a base). By the condition AB = BC = EA = 1.

We draw the height AH of the side face EAB (H belongs to EB), in a regular triangle AH will also be the median (BH = 1/2). CH – also the height of the side edge of the ECB. The CHA angle is the desired angle.

Let us find the lengths АH by the Pythagorean theorem:

AH = √ (AB² – BH²) = √ (1² – (1/2) ²) = √ (1 – 1/4) = √ (3/4) = √3 / 2.

The CH length will also be √3 / 2 (since the side faces are equal).

Let’s find the length АС, the diagonals of the square АBСD:

AC = √ (AB² + BC²) = √ (1² + 1²) = √2.

By the theorem of cosines in the triangle ACH:

AC² = AB² + BC² – 2 * AB * BC * cosH.

(√2) ² = (√ (3/4) ² + (√3 / 4) ² – 2 * √ (3/4) * √ (3/4) * cosH.

2 = 3/4 + 3/4 – 2 * 3/4 ​​* cosH.

2 = 6/4 – 3/2 * cosH.

2 = 1.5 – 1.5 * cosH.

1.5cosH = 1.5 – 2.

1.5cosH = -0.5.

cosH = -0.5: 1.5 = -1/3.

angle H = arccos (-1/3).

Answer: The angle is arccos (-1/3).