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# Find the angle between a plane and a line segment

- Question:
- EFGH-ABCD is a cube. GE is a diagonal in plane EFGH. Connect ED and FC. What is the degree measure of the angle between the plane EFCD and the line segment GE?

- Solution:
- Connect GB which intersects FC at point K, then connect EK.
- Because EF is perpendicular to FG and EF is perpendicular to FB, so EF is perpendicular to the plane FGCB.
- Because GB is on the plane FGCB, so EF is perpendicular to GB
- Because K in the intersection point of two diagonals FC and BG in square BCGF, so EF is perpendicular to GK.
- Because BCGF is a square, so its two diagonals are perpendicular and bisect each other, so GB is perpendicular to FC.
- Because K in the midpoint of GB, so GK is perpendicular to FC.
- Now we have, GK is perpendicular to EF and GK is perpendicular to FC.
- Because EF and FC are two intersected lines and two intersected lines make a plane, so GK is perpendicular to the plane EFCD.
- So the angle GEK is the angle between the line segment EG and the plane EFCD.
- In right triangle GEK, from sine definition,
- sin(angle GEK) = (opposite side of the angle GEK)/(hypotenuse of the right triangle GEK)
- = GK/GE = [(square root of 2)/2]a ÷ (square root of 2)a = 1/2
- So angle GEK = 30
^{o} - Therefore, the angle between the line segment EG and the plane EFCD is 30
^{o}.

If you are interested in watching the video for the solution of this problem, please click the video solution on how to find the angle between a line segment and a plane.