Through the vertex D of the obtuse angle of the rhombus ABCD, the perpendicular DM equal to 9.6 cm is drawn to its plane.

| July 17, 2021 | education

Through the vertex D of the obtuse angle of the rhombus ABCD, the perpendicular DM equal to 9.6 cm is drawn to its plane. The diagonals of the rhombus are 12 and 16 cm. Find the angle between the planes AMD and CDM.

Since the diagonals of the rhombus, at the point of intersection, are halved and intersect at right angles, then OA = AC / 2 = 16/2 = 8 cm, OD = BD / 2 = 12/2 = 6 cm.

In a right-angled triangle ADO, according to the Pythagorean theorem, we determine the length of the hypotenuse AD.

AD ^ 2 = AO ^ 2 + DO ^ 2 = 8 ^ 2 + 6 ^ 2 = 64 + 36 = 100.

AD = 10 cm.

The angle between the planes AMD and CDM is equal to the angle ADC of the rhombus, since AD and OD are perpendiculars to MD.

By the cosine theorem for the triangle AO ^ 2 = AD ^ 2 + CD ^ 2 – 2 * AD * CD * CosADC.

256 = 100 * 100 – 2 * 10 * 10 * CosADC.

200 * CosАDC = -256 + 200 = -56.

CosАDC = -56 / 200 = -0.28.

Angle ADC = Arcos (-0.28) = 106.260.

Answer: The angle between the planes is 106.260.



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