A circle is circumscribed about an isosceles triangle ABC with a base angle of 30 °

A circle is circumscribed about an isosceles triangle ABC with a base angle of 30 ° and a lateral side 10. Find the radius of the circle.

The radius of a circle circumscribed about an isosceles triangle is found from the formula:
R = a ^ 2 / 2h,
where R is the radius of the circumscribed circle, a is the lateral side of an isosceles triangle, h is the height of an isosceles triangle drawn to its base.
Draw the height BH in triangle ABC. BH divides ABC into two equal triangles ABH and BHC. Consider a triangle ABH. ABH – right-angled triangle (since BH is the height and is perpendicular to AC) with legs BH and AH and hypotenuse AB = 10 conventional units.
The BH leg lies against the angle at the base, which is 30 degrees. From the properties of a right-angled triangle, it is known that a leg lying opposite an angle of 30 degrees is equal to half of the hypotenuse. Then:
BH = AB / 2;
BH = 10/2;
BH = 5 conventional units.
Find the radius of the circumscribed circle:
R = 10 ^ 2/2 * 5 = 100/10 = 10 (conventional units).
Answer: R = 10 conventional units.