# The angle at the base of the isosceles triangle is 30 °, the lateral side is 16cm.

**The angle at the base of the isosceles triangle is 30 °, the lateral side is 16cm. Find the diameter of the circle around this triangle.**

Determine the value of the angle ABC.

In an isosceles triangle ABC, we draw the height of the ВН, which will also be the bisector and median. Then AH = AB * Cos30 = 16 * √3 / 2 = 8 * √3 cm.

Angle ABC = 180 – BAC – BCA = 180 – 30 – 30 = 120.

Then the degree measure of the larger arc AC = 2 * ABC = 2 * 120 = 240.

Then the degree measure of the arc ABC = 360 – 240 = 120.

The central angle AOC is equal to the degree measure of the arc ABC, the angle AOC = 120.

Then the triangle AOC is equilateral, since OA = OC = R, and the angle AOC = 120.

In an isosceles triangle AOC, we draw the height OH, which will also be the bisector and median. Angle ОАН = 180 – 90 – 60 = 30. Then AO = АН / Cos30 = 8 * √3 / (√3 / 2) = 16 cm.

Then the diameter of blood pressure = 2 * OA = 2 * 16 = 32 cm.

Answer: The diameter of the circle is 32 cm.