# A circle is inscribed in an isosceles trapezoid with an angle of 60, find the sides of the trapezoid if the length

**A circle is inscribed in an isosceles trapezoid with an angle of 60, find the sides of the trapezoid if the length of the segment that connects the points of contact of the circle with the lateral sides is 20 cm.**

We connect point O with points A, M and B. In a right-angled triangle МНО, the angle OMН = 30, the current as the angle BMН = ВAD = 60, and the angle OMВ = 90, as the radius to the tangent, then OMВ = 90 – 60 = 30.

The height of the PE divides the MK segment in half, then MH = MK / 2 = 20/2 = 10 cm.

ОМ = R = МН / Cos30 = 10 / (√3 / 2) = 20 / √3 cm.

Consider a right-angled triangle AOM. The AO segment, by the property of the tangent, divides the ВAD angle in half, then the OAM angle = 60/2 = 300.

Then AM = MO / tg30 = (20 / √3) / (1 / √3) = 20 cm.

By the property of a tangent drawn from one point, AE = AM = 20 cm, and since the trapezoid is isosceles, then DE = 20 cm.Then AD = 20 + 20 = 40 cm.

In a right-angled ВОМ triangle, the OBM angle = 120/2 = 60.

ВM = MO / / tg60 = (20 / √3) / √3 = 20/3 cm.

BP = BM = 20/3.

BC = 2 * BP = 40/3 = 13 (1/3) cm.

Then AB = СD = AM + BM = 20 + 20/3 = 80/3 = 26 (2/3) cm.

Answer: The sides of the trapezoid are equal: 26 (2/3) cm, 13 (1/3) cm, 26 (2/3) cm, 40 cm.