In triangle ABC, angle A is 10 degrees, angle C is 20 degrees, AC = 10cm.

In triangle ABC, angle A is 10 degrees, angle C is 20 degrees, AC = 10cm. Find the radius of the circle around this triangle.

We find the value of the angle B:

∠В = 180 ° – ∠А – ∠С = 180 ° – 10 ° – 20 ° = 170 ° – 20 ° = 150 °.

Using the sine theorem, we find the lengths of the sides AB and BC:

| BC | = | AC | * sin (A) / sin (B) = 10 * sin (10 °) / sin (150 °);

| AB | = | AC | * sin (C) / sin (B) = 10 * sin (20 °) / sin (150 °).

Find the area of ​​the triangle ABC on two sides and the angle between them:

S = | AC | * | AB | * sin (A) / 2 = 10 * (10 * sin (20 °) / sin (150 °)) * sin (10 °) / 2 = 50 * sin (10 °) * sin (20 °) / sin ( 150 °).

Using the formula for the area of ​​a triangle through the radius R of the circumscribed circle, we find R:

R = | AB | * | AC | * | BC | / (4 * S) = 10 * (10 * sin (10 °) / sin (150 °)) * (10 * sin (20 °) / sin (150 °)) / (4 * 50 * sin (10 ° ) * sin (20 °) / sin (150 °)) = (1000 * sin (10 °) * sin (20 °) / sin ^ 2 (150 °)) / (200 * sin (10 °) * sin ( 20 °) / sin (150 °)) = 5 / sin (150 °).

Answer: 5 / sin (150 °) cm.