A straight prism ABCDA1B1C1D1 is given. The base of the prism is a rhombus with a side of 8

| March 9, 2021 | education

A straight prism ABCDA1B1C1D1 is given. The base of the prism is a rhombus with a side of 8 and an acute angle of 45 degrees. The height of the prism is 6. Find the angle between plane AC1B and plane ABD.

Let’s construct the section AC1B, which will be the parallelogram AB1C1C.

From point D, construct a perpendicular to the edge AB of the rhombus ABCD.

The projection of the B1H segment onto the ABCD plane is the DH segment, then B1H is perpendicular to AB, then the linear angle DHB1 is the angle between the AC1B and ABD planes.

The ADN triangle is rectangular, then DH = AD * Cos45 = 8 * √2 / 2 = 4 * √2 cm.

From a right-angled triangle DD1Н tgDНD1 = DD1 / DH = 6/4 * √2 = 3/2 * √2 = 3 * √2 / 4.

Angle DD1H = arctan (3 * √2 / 4).

Answer: The angle between the planes is arctan (3 * √2 / 4).



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