Panjang sisi kubus ABCD.EFGH adalah alpha.beta adalah sudut

tan beta = sqrt(2)/2

Pembahasan

Untuk menentukan tan beta, di mana beta adalah sudut antara sisi FG dan bidang BGE pada kubus ABCD.EFGH dengan panjang sisi 'alpha', kita perlu melakukan analisis geometri:

Misalkan panjang sisi kubus adalah 'a' (menggantikan 'alpha' agar lebih umum).

1.  **Identifikasi Bidang BGE:** Bidang BGE dibentuk oleh titik B, G, dan E.
2.  **Proyeksikan Titik F ke Bidang BGE:** Sudut antara garis (sisi FG) dan bidang (BGE) adalah sudut antara garis FG dan garis proyeksinya pada bidang BGE. Garis proyeksi FG pada bidang BGE adalah GH (karena GH tegak lurus dengan BG dan GE dalam konteks kubus).
3.  **Segitiga Siku-siku:** Perhatikan segitiga siku-siku yang terbentuk. Salah satu cara adalah dengan mempertimbangkan segitiga yang dibentuk oleh titik F, proyeksinya pada bidang BGE, dan salah satu titik pada bidang BGE.

Alternatif lain yang lebih mudah:
Sudut antara FG (yang sejajar dengan BC dan EH) dan bidang BGE adalah sama dengan sudut antara EH dan bidang BGE.

Mari kita gunakan koordinat:
Misalkan A = (0, 0, 0), B = (a, 0, 0), C = (a, a, 0), D = (0, a, 0), E = (0, 0, a), F = (a, 0, a), G = (a, a, a), H = (0, a, a).

Sisi FG adalah garis dari F(a, 0, a) ke G(a, a, a). Vektor arah FG = G - F = (0, a, 0).

Bidang BGE melalui titik B(a, 0, 0), G(a, a, a), E(0, 0, a).

Untuk mencari sudut antara garis dan bidang, kita perlu vektor normal bidang.

Vektor BG = G - B = (0, a, a)
Vektor BE = E - B = (-a, 0, a)

Vektor normal bidang (N) = BG x BE
N =  | i   j   k |
    | 0   a   a |
    | -a  0   a |

N = i(a*a - a*0) - j(0*a - a*(-a)) + k(0*0 - a*(-a))
N = i(a^2) - j(a^2) + k(a^2)
N = (a^2, -a^2, a^2)
Kita bisa gunakan vektor normal yang disederhanakan: n = (1, -1, 1).

Sudut (theta) antara garis dengan vektor arah v dan bidang dengan vektor normal n diberikan oleh:
sin(theta) = |v . n| / (||v|| ||n||)

Di sini, v = (0, a, 0) (vektor arah FG)
n = (1, -1, 1)
v . n = (0)(1) + (a)(-1) + (0)(1) = -a
||v|| = sqrt(0^2 + a^2 + 0^2) = a
||n|| = sqrt(1^2 + (-1)^2 + 1^2) = sqrt(3)

sin(beta) = |-a| / (a * sqrt(3)) = a / (a * sqrt(3)) = 1/sqrt(3)

Jika sin(beta) = 1/sqrt(3), maka kita bisa mencari tan(beta) menggunakan identitas sin^2(beta) + cos^2(beta) = 1.
cos^2(beta) = 1 - sin^2(beta) = 1 - (1/sqrt(3))^2 = 1 - 1/3 = 2/3
cos(beta) = sqrt(2/3)

tan(beta) = sin(beta) / cos(beta) = (1/sqrt(3)) / (sqrt(2/3)) = (1/sqrt(3)) * (sqrt(3)/sqrt(2)) = 1/sqrt(2) = sqrt(2)/2.

Namun, perlu diperiksa lagi sudut beta yang dimaksud. Jika beta adalah sudut antara sisi FG dan bidang BGE, maka proyeksi F ke bidang BGE adalah H. Jadi kita melihat sudut FGH. Dalam kubus, FG tegak lurus GH. Ini berarti sudutnya 90 derajat, yang tidak masuk akal untuk tan beta. Kemungkinan interpretasi lain adalah sudut antara garis FG dan proyeksinya pada bidang BGE.

Mari kita tinjau ulang geometri kubus. Sisi FG sejajar dengan EH. Kita cari sudut antara EH dan bidang BGE.

Perhatikan bidang EFGH. Titik E, H, G berada di bidang ini. Titik B juga merupakan bagian dari bidang BGE.

Dalam kubus, sisi FG sejajar dengan sisi EH. Sudut antara garis FG dan bidang BGE sama dengan sudut antara garis EH dan bidang BGE.

Proyeksi titik E pada bidang BGE adalah E itu sendiri. Proyeksi titik H pada bidang BGE adalah titik H itu sendiri. Ini tidak membantu.

Mari kita cari sudut antara garis yang 'melalui' FG dan memotong bidang BGE. Garis FG tegak lurus dengan bidang ABFE dan DCGH. Bidang BGE memotong bidang DCGH sepanjang garis GH.

Consider the triangle FGH. FG = a, GH = a. FH is the diagonal of the square face EFGH, FH = a√2.

Let's reconsider the angle definition. The angle between a line and a plane is the angle between the line and its projection onto the plane. The projection of the line segment FG onto the plane BGE needs to be determined.

Let's use a different approach. Consider the right-angled triangle formed by F, G, and the intersection of the perpendicular from F to plane BGE. 

Alternatively, consider the angle between FG and BG, and FG and GE. This is not directly the angle with the plane.

Let's use the property that the angle between line FG and plane BGE is the angle between FG and its projection onto plane BGE. 

Consider the plane containing FG and perpendicular to BGE. This seems complicated.

Let's re-evaluate the projection of F onto plane BGE. The line FG is parallel to the line EH. The angle between line FG and plane BGE is the same as the angle between line EH and plane BGE.

Let's consider the distance from F to the plane BGE. Let this distance be 'd'. The angle beta is such that sin(beta) = d / length(FG).

Consider the triangle BGE. BG = a√2, BE = a√2, GE = a√2. Triangle BGE is equilateral.

Let's go back to coordinates and vector normal.
n = (1, -1, 1).
Line FG direction vector v = (0, a, 0).
Angle phi between v and n is given by cos(phi) = (v . n) / (||v|| ||n||) = (-a) / (a * sqrt(3)) = -1/sqrt(3).
This is the angle between the line and the normal to the plane. The angle beta between the line and the plane is 90 - phi.

sin(beta) = cos(phi) = |-1/sqrt(3)| = 1/sqrt(3).

From sin(beta) = 1/sqrt(3), we found cos(beta) = sqrt(2/3), and tan(beta) = 1/sqrt(2) = sqrt(2)/2.

Let's verify if this interpretation of beta is correct. The angle between a line and a plane is defined as the smallest angle between the line and any line lying in the plane. This angle is formed by the line and its orthogonal projection onto the plane.

Let's consider the projection of FG onto the plane BGE. Since FG is parallel to EH, we can consider the projection of EH onto the plane BGE. However, E and H are in the plane, which is incorrect. 

Consider the face BCGF. FG is perpendicular to BG in this face. This is incorrect.

Let's assume alpha refers to the length of the side.

Consider the cube again. FG is an edge. BGE is a plane. The angle beta is between the edge FG and the plane BGE.

Consider the plane passing through F and perpendicular to BGE. This is too complex.

Let's reconsider the vector approach. FG direction is (0, 1, 0) if we align F at (0,0,0) and G at (0,1,0) in a local system for that edge. This is not helpful.

Let's stick to the coordinate system A=(0,0,0) ... G=(a,a,a).
FG vector = (0, a, 0).
Plane BGE normal vector n = (1, -1, 1).

Angle between FG and plane BGE is beta.
sin(beta) = |FG . n| / (||FG|| ||n||) = |(0,a,0) . (1,-1,1)| / (a * sqrt(1^2 + (-1)^2 + 1^2))
sin(beta) = |0 - a + 0| / (a * sqrt(3)) = |-a| / (a * sqrt(3)) = a / (a * sqrt(3)) = 1/sqrt(3).

If sin(beta) = 1/sqrt(3), then tan(beta) = sin(beta) / cos(beta).
cos(beta) = sqrt(1 - sin^2(beta)) = sqrt(1 - (1/sqrt(3))^2) = sqrt(1 - 1/3) = sqrt(2/3).
tan(beta) = (1/sqrt(3)) / (sqrt(2/3)) = (1/sqrt(3)) * (sqrt(3)/sqrt(2)) = 1/sqrt(2) = sqrt(2)/2.

The question uses 'alpha.beta' which seems like a typo and likely means 'alpha' is the side length, and 'beta' is the angle. Assuming 'alpha' is the side length of the cube.

Therefore, tan beta = sqrt(2)/2.