In triangle ABC, ∠BAC = 38, ∠ACB = 22. Find the radius of the circle circumscribed about this triangle

In triangle ABC, ∠BAC = 38, ∠ACB = 22. Find the radius of the circle circumscribed about this triangle if the distance from its center to the AC side of the triangle is 12.5.

Decision:

angle ABC = 180 – (38 + 22) = 120 degrees

Angle ABC – inscribed, which means it is equal to half of the arc ADC. Hence:

arc ADC = 2 * angle ABC = 2 * 120 = 240 degrees

arc ABC = 360 – arc ADC = 360 – 240 = 120 degrees

The angle AOC is central, which means it is equal to the arc ABC, i.e. 120 degrees.

ОА = ОС = R, hence the triangle AОС is isosceles and the height ОН in it is a bisector.

angle AOH = 1/2 * angle AOC = 120: 2 = 60 degrees

angle OAH = 90 – 60 = 30 degrees

OH = 1/2 * AO

AO = 2OH = 2 * 12.5 = 25

Answer: R = 25.