The quadrilateral ABCD is inscribed in a circle. Beams AB and DC meet at point K

The quadrilateral ABCD is inscribed in a circle. Beams AB and DC meet at point K, and diagonals AC and BD meet at point N. Angle BNC is 68 and angle AKD is 36. Find the angle BAC.

The internal angle BNC, formed by the intersection of the diagonals AC and VD, is equal to the half-sum of the degree measures of the arcs on which this and the opposite vertical angle rests.

BNC angle = (AD + ⌒BC) / 2 = 680.

⌒AD = 136 – ⌒BC. (1).

The AKD angle is equal to the half-difference of the degree measures of the arcs formed by the secant lines AK and DC.

Angle ADK = (⌒AD – ⌒BC) / 2 = 360.

⌒AD = 72 + ⌒BC. (2).

Let’s equate expressions 1 and 2.

136 – BC = 72 + ⌒BC.

2 * BC = 136 – 72 = 64.

Arc BC = 64/2 = 32.

The inscribed angle BAC rests on the arc BC, then the angle BAC = 32/2 = 160.

Answer: The value of the angle BAC is 160.