In triangle ABC AC = 10, angle BAC = 70 degrees, angle ACB = 80 degrees. Find the radius of the circle around this triangle
In solving the problem, we will use the theorem of sines. To do this, we find the angle ABC, located opposite the side of the AC.
∠ ABC = 180 ° – (∠ BAC + ∠ ACB) = 180 ° – 150 ° = 30 °.
By the sine theorem, we write the equality:
2R = AC / sin 30 ° = 10 / 1/2 = 20.
R = 10.
Answer: the radius of the circumscribed circle is 10.
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