The sum of the outer angles of triangle ABC at vertices A and B, taken one for each vertex

The sum of the outer angles of triangle ABC at vertices A and B, taken one for each vertex, is 240 degrees. Find corner C.

The outer corner for a corner of a triangle is its adjacent corner.

Let ABC ∠MAC, ∠НBC, and ∠KCB be external corners for ∠A, ∠B, and ∠C, respectively.

1. Since the sum of adjacent angles is 180 °, then:

∠MAC = 180 ° – ∠A;

∠НBC = 180 ° – ∠B;

∠KCB = 180 ° – ∠C.

By condition, the sum of the outer angles at the vertices A and B is 240 °, then:

∠MAC + ∠НBC = 240 °.

Let’s replace:

180 ° – ∠A + 180 ° – ∠B = 240 °;

– ∠A – ∠B = 240 ° – 360 °;

– (∠A + ∠B) = – 120 °;

∠A + ∠B = 120 °.

2. By the theorem on the sum of the angles of a triangle, in △ ABC:

∠A + ∠B + ∠C = 180 °;

120 ° + ∠C = 180 °;

∠C = 180 ° – 120 °;

∠C = 60 °.

Answer: ∠C = 60 °.