Question

Равнобокая трапеция вписана в окружность центр которой принадлежит одному из оснований. угол между диагоналями трапеции противолежащей её боковой стороне равен 56 градусов . найдите углы трапеции. ПОЖАЛУЙСТА СРОЧНО!!!

Answer 1

Вертикальный данному углу тоже 56
а смежные по (180-56=124) 124° каждый
если я правильно поняла задачу

Answer 2

Problem Analysis

We are given that a trapezoid is inscribed in a circle, with the center of the circle belonging to one of the bases. We are also given that the angle between the diagonals of the trapezoid and the opposite side is 56 degrees. We need to find the angles of the trapezoid.

Solution

To find the angles of the trapezoid, we can use the properties of inscribed angles and the fact that the sum of the angles in a trapezoid is 360 degrees.

Let's denote the trapezoid as ABCD, with AB and CD as the bases, and AC and BD as the diagonals. Let O be the center of the circle.

1. Angle AOB is a central angle and is equal to twice the angle between the diagonals (56 degrees). Therefore, angle AOB = 2 * 56 = 112 degrees.

2. Angle ACB is an inscribed angle that intercepts the same arc as angle AOB. Therefore, angle ACB = 112 degrees.

3. Angle BCD is a base angle of the trapezoid. Since the sum of the angles in a trapezoid is 360 degrees, we can find angle BCD by subtracting the other angles from 360 degrees. Angle BCD = 360 - angle ACB = 360 - 112 = 248 degrees.

4. Angle ADB is an inscribed angle that intercepts the same arc as angle BCD. Therefore, angle ADB = angle BCD = 248 degrees.

Therefore, the angles of the trapezoid are: - Angle ACB = 112 degrees - Angle BCD = 248 degrees - Angle ADB = 248 degrees

Answer

The angles of the trapezoid are: - Angle ACB = 112 degrees - Angle BCD = 248 degrees - Angle ADB = 248 degrees