One angle of the triangular is 10 degrees more than the second and 20 degrees more than the third

One angle of the triangular is 10 degrees more than the second and 20 degrees more than the third angle. Find the corners of the triangle.

Let the angle ABC = x, then the angle BCA = x – 10 ° (since it is 10 ° less than ABC).

Angle BAC = x – 20 ° (since it is less than ABC by 20 °).

It is known that the sum of the interior angles of a triangle = 180 °.

(ABC + BAC + BCA = 180).

We get the equation: x + x – 10 ° + x – 20 ° = 180 ° or 3 * x = 210 °.

We find the value of x and the angle ABC: x = 70 °.

Find the remaining angles: BCA = x – 10 ° = 70 ° – 10 ° = 60 °.

BAC = x – 20 ° = 70 ° – 20 ° = 50 °.

Answer: The angles of the triangle are 70 °, 60 ° and 50 °.