On the continuation of side AB of triangle ABC beyond point A, point K is taken so that AK = AC
On the continuation of side AB of triangle ABC beyond point A, point K is taken so that AK = AC, and on its continuation beyond point B, point M is taken so that BM = BC. Find the angles of the triplet MKC if the angle BAC = 70, ABC = 80 degrees.
In triangle ABC, angle BAC is 70 °, angle ABC is 80 °. Find the angle of the BCA (the sum of the angles in the triangle is 180 °):
180 ° – (70 ° + 80 °) = 30 °.
Let AC = x, then KC will be equal to 2x (AK = AC). Let BC = y, then MC will be equal to 2y (BM = BC).
Consider triangles ABC and ISS: angle C is common; AC / KC = x / 2x = 1/2; BC / MC = y / 2y = 1/2.
Consequently, triangles are similar in two sides and the angle between them.
For similar triangles, the corresponding angles are equal.
Hence, the angle K is equal to the angle A = 70 °, the angle M is equal to the angle B = 80 °, the angle C is equal to 30 °.