Angle B of triangle ABC is twice less than angle A and 20 degrees more than angle C. Find the angles of the triangle

Given: ∠А = 2 * ∠В, ∠ В = ∠С + 20 °. We have: 2 * (∠С + 20 °) + ∠С + 20 ° + ∠С = 180 ° or 4 * ∠С = 120 °, whence ∠С = 120 ° / 4 = 30 °, ∠В = ∠С + 20 ° = 30 ° + 20 ° = 50 °, ∠А = 2 * ∠В = 2 * 50 ° = 100 °.
Answer: ∠А = 100 °, ∠В = 50 °, ∠С = 30 °.



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