The lateral edge of a regular triangular pyramid is inclined to the base plane at an angle of 30.

The lateral edge of a regular triangular pyramid is inclined to the base plane at an angle of 30. And at what angle to the plane of the base is its lateral face tilted?

Let the length of the height SO = X cm.

Then, in a right-angled triangle SOB tg30 = X / OB.

OB = X / tg30 = X / (1 / √3) = X * √3 cm.

The median BK, at point O, is divided in the ratio 2/1, then KO = OB / 2 = X * √3 / 2 cm.

Since the ABC triangle is equilateral, OH = OK = X * √3 / 2 cm.

In a right-angled triangle SOH, tgSHO = SO / OH = X / (X * √3 / 2) = 2 / √3.

Angle SHO = arctan (2 / √3) ≈ 490.

Answer: The angle between the side face and the plane of the base is 490.