In a regular triangular pyramid, the side of the base is 2√3 cm, and the height is 2 cm.

In a regular triangular pyramid, the side of the base is 2√3 cm, and the height is 2 cm. Find the angle of inclination of the side edge to the plane of the base.

SABC – regular triangular pyramid, SO = 2 cm – height, AB = BC = AC = 2√3 cm.
1. Since SABC is a regular triangular pyramid, a regular (equilateral triangle) lies at its base. SO – falls into the center of the circle described around ABC. Let us find the length of AO (CO, BO) – the radius of the circumscribed circle.
The radius of the circumscribed circle of a regular triangle:
R = √3 / 3 * a,
where a is the side length of a regular triangle.
R = √3 * 2√3 / 3 = 2 * 3/3 = 2 (cm).
2. Consider a triangle SOC: angle SOC = 90 degrees, since SO is height, then SOC is a right-angled triangle, and SO = 2 cm and OC = 2 cm are legs, SC is hypotenuse. Since SO = 2 cm and OS = 2 cm, SO = OS = 2 cm, therefore SOС is an isosceles right triangle, SC is the base, the angles SСО and СSO are the angles at the base, which are equal. We denote the angles SCO and CCO as x, then by the theorem on the sum of the angles of a triangle:
angle SСО + angle СSO + angle SОС = 180 degrees;
x + x + 90 = 180;
2x = 180 – 90;
2x = 90;
x = 90/2;
x = 45 degrees.
SCO angle = CCO angle = x = 45 degrees.
Answer: 45 degrees.