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# How to find the angle between a line and a plane in regular tetrahedron?

Question:

In the figure shown, P-ABC is a regular tetrahedron with side length a. Find the angle between line segment PA and the plane ABC. Solution:

Because P-ABC is a regular tetrahedron with side length a, PA = PB = PC = AB = BC = CA = a. Each triangle is an equilateral triangle with side length a.

Draw a line from the vertex P perpendicular to plane ABC, this line intersects the plane ABC at the point O. Connect OA. OA is the projection of PA on the plane ABC. So, the angle between PA and the plane ABC is the angle between PA and OA.

Because PO perpendicular to OA, triangle POA is a right triangle.

Let the angle PAO is the angle theta. In right triangle POA, cos theta = OA/PA

Because P-ABC is a regular tetrahedron, the point O is the center of the equilateral triangle ABC. AO is the distance from the vertex A to the center of the equilateral triangle. Please see the property of the equilateral triangle. AO = (square root of 3) a/3 in which a is the side of the equilateral triangle.

cos theta = OA/PA
= (square root of 3) a/3 ÷ a
= (square root of 3)/3
= 0.5774
angle theta = arc cos 0.5774 = 54.73 degrees

Therefore, the angle between PA ang the plane ABC is 54.73 degrees. Please watch the video for more details.