Question

Угол между диагональю прямоугольника и одной стеной составляет 30 ° и радиус окружности внешней диагонали R. Найдите наименьшую стенку прямоугольника

Answer 1

Пусть АС -диагональ прямоугольника АВСД, она является диаметром описанной окружности, и одновременно -гипотенузой  треугольника АСД. Если угол САД равен 30°, то СД (меньшая сторона) равно половине АС, т.е. R.

Answer 2

I will try to answer your question in detail. The angle between the diagonal of a rectangle and one wall is 30 ° and the radius of the circle of the outer diagonal is R. Find the smallest wall of the rectangle.

Let us denote the length and width of the rectangle by a and b, respectively. Then, the diagonal of the rectangle is d, and the angle between the diagonal and the wall is α. We are given that α = 30° and R = d/2. We need to find the value of b, which is the smallest wall of the rectangle.

We can use the following formulas to solve this problem:

- The formula for the diagonal of a rectangle through the length and width: d = √(a^2 + b^2) - The formula for the side of a rectangle (length and width) through the diagonal and the angle α: a = d sinα, b = d cosα

Substituting the given values, we get:

d = 2R

a = 2R sin 30° = R

b = 2R cos 30° = R√3

Therefore, the smallest wall of the rectangle is b = R√3.