Calculate the external angles: a right-angled triangle, one of the corners of which is 40 degrees.

Given a rectangular △ ABC: ∠C = 90 °, ∠A = 40 °.
1. By the theorem on the sum of the angles of a triangle:
∠A + ∠B + ∠C = 180 °;
40 ° + ∠B + 90 ° = 180 °;
∠B = 180 ° – 130 °;
∠B = 50 °.
2. The outer angle of one of the angles of the triangle is an adjacent angle, and the sum of the adjacent angles is 180 °.
∠MCA – outside angle at vertex C, ∠KAB – outside angle at vertex A, ∠FBA – outside angle at vertex B.
Thus:
a) ∠MCA + ∠C = 180 °;
∠MCA + 90 ° = 180 °;
∠MCA = 180 ° – 90 °;
∠MCA = 90 °;
b) ∠KAB + ∠A = 180 °;
∠KAB + 40 ° = 180 °;
∠KAB = 180 ° – 40 °;
∠KAB = 140 °;
c) ∠FBA + ∠B = 180 °;
∠FBA + 50 ° = 180 °;
∠FBA = 180 ° – 50 °;
∠FBA = 130 °.
Answer: ∠KAB = 140 °, ∠FBA = 130 °, ∠MCA = 90 °.