In triangle ABC, bisectors Am and BN are drawn, intersecting at point K, and the angle AKN = 58gr. Find: angle ACB.

Let’s denote the angles of the triangle as follows:

angle A = x;

angle B = y;

angle C = z;

Then we get that the following equality is true:

x + y + z = 180 °;

Since we know from the condition that the AKN angle is 58 °, the BKA angle will be 180 ° – 58 ° = 122 °.

Let’s take a closer look at the AKB triangle:

angle AKB = 122 °;

angle KAB = 1/2 angle BAN = x / 2;

angle KBA = 1/2 angle CBA = y / 2.

Then we get, which is true for the triangle BKA:

angle AKB + angle KAB + angle KBA = 180 °;

122 ° + x / 2 + y / 2 = 180 °;

x / 2 + y / 2 = 180 ° – 122 °;

(x + y) / 2 = 58 °;

x + y = 58 ° * 2 = 116 °.

We substitute this value into the expression with three unknowns and we get:

116 ° + z = 180 °;

z = 180 ° – 116 °;

z = 64 °.

Answer: angle ACB = 64 °.