# 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.

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