Point A and B divide the circle with center O into arcs AMB and ACB so that the arc ACB is 60 degrees less than the AMB arc. AM is the diameter of the circle. Find the angles AMB, ABM, ACB.
The sum of the degree measures of arcs ACB + AMB = 360.
By condition, the arc AMD = ACB + 60.
Then ACB + 60 + ACB = 360.
2 * ACB = 360 – 60 = 300.
Arc ACB = 300/2 = 150.
Then the arc AMB = 150 + 60 = 210.
The inscribed angle ACB rests on the AMB arc, then the ACB angle is equal to half the degree measure of the AMB arc.
Angle ACB = 210/2 = 105.
The ABM triangle is rectangular, since the inscribed ABM angle rests on the diameter of the circle, then the ABM angle = 180/2 = 90.
The inscribed angle АМВ rests on the arc АСВ, then the angle АМВ = АСВ / 2 = 150/2 = 75.
Answer: Angle AMB = 75, angle ABM = 90, angle ACB = 105.
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