In an acute-angled triangle ABC, the height BH and the bisector AD are drawn, intersecting at exactly O.

In an acute-angled triangle ABC, the height BH and the bisector AD are drawn, intersecting at exactly O. It turned out that the angle AOB is 4 times greater than the angle DAB. What is the CAB angle?

Let the sought angle CAB be equal to 2x, then the angle BAD is equal to x, since AD is the bisector of the CAB angle, and the AOB angle is 4x by condition. In a right-angled triangle ВНA, the AВН angle is 180 ° – 90 ° – 2x = 90 ° – 2x. Thus, in triangle AOB, the sum of its three angles is

x + 4x + 90 ° – 2x = 180 °.

From where you can calculate the unknown value x.

For what we will give similar ones:

3x = 90 °.

Divide both sides of the equation by 3:

x = 30 °.

As a result, the CAB angle is 2 * 30 ° = 60 °.

Answer: the CAB angle is 60 °.