In the triangle ABC, the heights AH and BT are drawn, O is the point of their intersection.

In the triangle ABC, the heights AH and BT are drawn, O is the point of their intersection. Find AOB if CAB = 48gr, CBA = 53g.

In the triangle ABC, we determine the value of the angle ACB. The sum of the inner angles of the triangle is 180, then the angle ACB = (180 – ABC – BAC) = (180 – 53 – 48) = 79.

Since AH and BT are the heights of the ABC triangle, the angle AНС = BTC = 90.

The sum of the inner angles of the quadrilateral CHOT is 360, then the angle of the CHOT = (360 – 80 – 90 – 79) = 101.

Angle AOB = HOT = 101 as vertical angles at the intersection of heights BT and AН.

Answer: Angle AOB is 101.