The bisectors of triangle ABC intersect at point O. Find the angle A of this triangle if the angle BOC = 100 degrees.

Consider a triangle COB: angle O = 100 °, so the sum of the angles OBC and OCB is equal to:

ОBС + ОСB = 180 ° – BОС = 180 ° – 100 ° = 80 °.

Since BО is a bisector, then the angle B of the triangle ABC is equal to two angles OBC: B = 2 * OBC.

Since CO is a bisector, the angle C of the triangle ABC is equal to two angles of the OCB: C = 2 * OCB.

Let us express the sum of the angles B and C:

B + C = 2 * OBC + 2 * OCB = 2 (OBC + OCB) = 2 * 80 ° = 160 °.

The sum of the angles in the triangle is 180 °, so the angle A is equal to:

A = 180 ° – (B + C) = 180 ° – 160 ° = 20 °.

Answer: Angle A is 20 °.