In a circle centered at point O, diameters AD and BC are drawn, the angle OAB is 29 degrees

In a circle centered at point O, diameters AD and BC are drawn, the angle OAB is 29 degrees, find the value of the angle BDA.

Let us construct a circle with the center O. Through the center O we draw the diameter AD and BC.

Consider a triangle AOB and COD-equal, because AO = BO = CO = DO = R, angles AOB and DOC-vertical and equal. Angle ОАВ = ОDC = 29 ° – intersecting, they rest on equal arcs ВD = AC = 2 * angle ОАВ = 2 * 29 = 58 °.

ВD + АС = 58 + 58 = 116 °.

Arc CD = AB (mk angle COD = AOB-central, rest on these arcs):

AB = (360 ° – 116 °) / 2 = 122 °.

Angle BDA rests on arc AB:

BDA = 122 ° / 2 = 61 °