In an acute-angled triangle ABC, the heights of AA1 and BB1 intersect at point O, a) prove that angle BAO

In an acute-angled triangle ABC, the heights of AA1 and BB1 intersect at point O, a) prove that angle BAO = angle BCO b) find the angles of triangle ABC if angle BCO = 28 degrees, and angle ABB1 = 44 degrees

Let’s build the height CC1.

By the property of the heights of the triangle, they intersect at one point, then CC1 passes through point O.

In right-angled triangles AA1B and CC1B, angle B is common.

The sum of the angles (ABA1 + A1AB) = (CBC1 + C1CB) = 90.

Since the angles ABA1 = CBC1, then the angle A1AB = C1CB, which was required to prove.

In a right-angled triangle ABB1, we determine the value of the angle BAB1.

Angle BAB1 = (90 – 44) = 46.

We have proved that angle ABA1 = BCC1, angle BAA1 = 28, then angle A1AC = BAA1 – BAA1 = 46 – 28 = 18.

Then in a right-angled triangle АА1С the angle АСА1 = 90 – 18 = 72.

Then the angle ABC = (180 – 46 – 72) = 62.

Answer: The angles of the triangle ABC are equal to 46, 72, 62.