The bisector of angle A of the parallelogram ABKC intersects the side BC at point M.
| May 8, 2021 | education
The bisector of angle A of the parallelogram ABKC intersects the side BC at point M. Find the angles of the parallelogram if the angle BMK is known to be 146 degrees.
Let the angle BAK = X0, then the angle ABC = (180 – X) 0, the angle BAM = X / 2.
The angle AMC is the outer angle of the triangle ABM, then the angle AMC = (ABM + BAM) = (180 – X + X / 2) = 146.
0.5 * X = (180 – 146) = 34.
X = VAK = 34 / 0.5 = 68, then the angle ABC = (180 – 68) = 112.
Answer: The angles of the parallelogram are 68 and 112.
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