In the rhombus ABCD, the angle BAC intersects the side BC and the diagonal BD

| April 12, 2021 | education

In the rhombus ABCD, the angle BAC intersects the side BC and the diagonal BD, respectively, at points M and N, the angle AMC = 120 degrees. find the angle value ANB.

Let the value of the acute angle of the rhombus be X0, the angle BAD = X0, then the angle BAO = X / 2.

The angle AMB and AMC are adjacent angles, the sum of which is 180, then the angle AMB = 180 – 120 = 60.

The sum of the adjacent angles of the rhombus is 180, then the angle ABC = 180 – ABD = 180 – X.

The diagonals of the rhombus are the bisectors of the angles at the vertices of the rhombus, then the angle BAO = BAD / 2 = X / 2.

AM is the bisector of the angle BAC, then the angle BAM = BAO / 2 = (X / 2) / 2 = X / 4.

In triangle ABM, the sum of the outer angles will be. X / 4 + (180 – X) + 60 = 180.

3 * X / 4 = 60.

X = BAD = 60 * 4/3 = 80.

Then the angle ABC = 180 – 80 = 100.

ABO angle = 100/2 = 50.

In triangle ABN, angle ANB = 180 – BAN – ABN = 180 – 20 – 50 = 110.

Answer: Angle ANB is 110.



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