A quadrilateral ABCD is inscribed in a circle. Angle A is 58 degrees greater than angle B

A quadrilateral AВСD is inscribed in a circle. Angle A is 58 degrees greater than angle B and 4 times greater than angle C. Find the corners.

Decision:

Let the angle C be equal to x, then the angle A is equal to 4x.

By the property of the inscribed quadrilateral:

angle A + angle C = angle B + angle D = 180 degrees.

We get the equation:

x + 4x = 180

5x = 180

x = 180: 5

x = 36

angle C = 36 degrees

angle A = 36 x 4 = 144 degrees

angle B = angle A – 58 = 144 – 58 = 86 degrees

angle D = 180 – angle B = 180 – 86 = 94 degrees

Answer: angle A = 144 degrees, angle B = 86 degrees, angle C = 36 degrees, angle D = 94 degrees.