The base side of a regular prism ABCA1B1C1 is 2 and the side edge is √71 Find the sine of the angle

The base side of a regular prism ABCA1B1C1 is 2 and the side edge is √71 Find the sine of the angle between line CB1 and plane AA1C1.

Since the prism is regular, there are equilateral triangles at its bases.

From the vertex B1 we construct the height B1H, which is also the median of the triangle.

Then C1H = A1C1 / 2 = 2/2 = 1 cm.

The triangle CC1H is rectangular, then, according to the Pythagorean theorem, CH ^ 2 = CC1 ^ 2 + C1H ^ 2 = 71 + 1 = 72.

CH = √72 cm.

In a right-angled triangle BCB1, according to the Pythagorean theorem, CB1 ^ 2 = BB1 ^ 2 + BC ^ 2 = 71 + 4 = 75.

CB1 = √75 cm.

Plane А1В1С1 is perpendicular to plane АА1С1С, then В1Н is perpendicular to СН, and triangle СНВ1 is rectangular.

CosHCB1 = CH / CB1 = √72 / √75 = 8.48 / 8.66 = 0.98.

Angle НСВ1 = arcos0.98 ≈ 11.5.

Answer: The angle between a straight line and a plane is 11.5.



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