T ABCD adalah limas beraturan dengan semua rusuknya sama

Kosinus sudut antara bidang TAD dan TBC adalah 1/3.

Pembahasan

Karena limas T ABCD beraturan dengan semua rusuknya sama panjang, maka alas ABCD adalah persegi dan sisi-sisi tegaknya adalah segitiga sama sisi. Misalkan panjang rusuknya adalah 'a'.

Untuk mencari kosinus sudut antara bidang TAD dan TBC, kita perlu mencari garis potong kedua bidang tersebut dan mengambil sebuah titik pada garis potong tersebut. Misalkan titik potong diagonal alas AC dan BD adalah O.

Karena limas beraturan, TO tegak lurus bidang ABCD.

Bidang TAD dan TBC berpotongan pada garis yang melalui T dan sejajar AD dan BC. Namun, untuk menentukan sudut antara dua bidang, lebih mudah jika kita menggunakan proyeksi.

Kita dapat memproyeksikan bidang TAD ke bidang ABCD. Proyeksi bidang TAD adalah garis AD. Proyeksi bidang TBC adalah garis BC.

Sudut antara bidang TAD dan TBC sama dengan sudut antara garis TO (tinggi limas) dan proyeksinya pada bidang alas, namun ini tidak tepat karena kita perlu mencari garis yang tegak lurus dengan garis potong kedua bidang tersebut.

Alternatif lain adalah dengan memilih titik pada alas dan menarik garis tegak lurus ke kedua bidang.

Mari kita gunakan pendekatan vektor atau proyeksi yang lebih tepat.

Karena ABCD adalah persegi dengan rusuk 'a', maka jarak antara AD dan BC adalah 'a'.

Kita bisa mengambil titik tengah rusuk AD, sebut saja P, dan titik tengah rusuk BC, sebut saja Q. Maka PQ tegak lurus AD dan BC, dan PQ = a.

Sekarang kita perhatikan segitiga TPQ. TO adalah tinggi limas, dan O adalah titik tengah PQ. TP dan TQ adalah tinggi segitiga sama sisi TAD dan TBC. Jadi, TP = TQ = (a√3)/2.

Dalam segitiga sama sisi dengan rusuk 'a', tingginya adalah (a√3)/2.

Segitiga TPQ adalah segitiga sama kaki dengan TP = TQ.

Untuk mencari sudut antara bidang TAD dan TBC, kita perlu mencari garis yang tegak lurus terhadap garis potong kedua bidang tersebut.

Jika kita memotong limas dengan bidang yang tegak lurus terhadap alas dan membagi dua sudut dihedral antara TAD dan TBC, maka kita akan mendapatkan sudut yang kita cari.

Consider a plane passing through T and perpendicular to AD and BC. Let this plane intersect AD at P and BC at Q. Since the pyramid is regular, P and Q are midpoints of AD and BC respectively. Thus, PQ is parallel to AB and CD, and PQ = a.
Also, TP and TQ are altitudes of the equilateral triangles TAD and TBC, so TP = TQ = (a√3)/2.
The line of intersection of planes TAD and TBC is a line through T parallel to AD and BC. Let's call this line L.

A better approach is to use the dihedral angle definition.

Let's consider the angle between the planes TAD and TBC. Since the pyramid is regular and all edges are equal, the base ABCD is a square, and the faces TAD and TBC are equilateral triangles.

Let M be the midpoint of AD and N be the midpoint of BC. Then TM is perpendicular to AD and TN is perpendicular to BC. The length of TM and TN is the height of the equilateral triangle, which is (a√3)/2.
The line segment MN is perpendicular to both AD and BC and MN = a.
The angle between the planes TAD and TBC is the angle between TM and TN, if they were intersecting perpendicularly to the line of intersection. This is not the case.

The line of intersection of the planes TAD and TBC is a line through T parallel to AD and BC. Let's call this line L.

Let's reconsider the geometry. Let the vertices of the base be A=(-a/2, -a/2, 0), B=(a/2, -a/2, 0), C=(a/2, a/2, 0), D=(-a/2, a/2, 0). Let the apex be T=(0, 0, h). Since all edges are equal, TA=TB=TC=TD=a.
TA^2 = (0 - (-a/2))^2 + (0 - (-a/2))^2 + (h - 0)^2 = (a/2)^2 + (a/2)^2 + h^2 = a^2/4 + a^2/4 + h^2 = a^2/2 + h^2.
Since TA = a, a^2 = a^2/2 + h^2, which means h^2 = a^2/2, so h = a/√2.

Now consider the plane TAD. A normal vector to this plane can be found. The vectors AD = (0, a, 0) and AT = (-a/2, -a/2, a/√2).
A normal vector n1 is AD x AT = (0, a, 0) x (-a/2, -a/2, a/√2) = (a(a/√2) - 0, 0 - 0, 0 - (-a/2)a) = (a^2/√2, 0, a^2/2).
We can simplify this normal vector to (√2, 0, 1).

Now consider the plane TBC. The vectors BC = (0, a, 0) and BT = (-a/2, a/2, a/√2).
A normal vector n2 is BC x BT = (0, a, 0) x (-a/2, a/2, a/√2) = (a(a/√2) - 0, 0 - 0, 0 - (-a/2)a) = (a^2/√2, 0, a^2/2).
Wait, the normal vectors are the same. This indicates the planes are parallel or identical, which is incorrect for adjacent faces.

Let's redefine coordinates for simplicity. Let A=(0,0,0), B=(a,0,0), C=(a,a,0), D=(0,a,0). The center of the base is (a/2, a/2, 0). Since all edges are equal, T must be directly above the center of the base. So T = (a/2, a/2, h).
TA^2 = (a/2 - 0)^2 + (a/2 - 0)^2 + (h - 0)^2 = a^2/4 + a^2/4 + h^2 = a^2/2 + h^2.
Since TA = a, a^2 = a^2/2 + h^2, so h = a/√2.
T = (a/2, a/2, a/√2).

Plane TAD: Passes through A=(0,0,0), D=(0,a,0), T=(a/2, a/2, a/√2).
Vector AD = (0,a,0). Vector AT = (a/2, a/2, a/√2).
Normal vector n1 = AD x AT = (0,a,0) x (a/2, a/2, a/√2) = (a(a/√2), 0, -a(a/2)) = (a^2/√2, 0, -a^2/2).
We can simplify n1 to (√2, 0, -1).
The equation of plane TAD is √2x - z = 0.

Plane TBC: Passes through B=(a,0,0), C=(a,a,0), T=(a/2, a/2, a/√2).
Vector BC = (0,a,0). Vector BT = (a/2 - a, a/2 - 0, a/√2 - 0) = (-a/2, a/2, a/√2).
Normal vector n2 = BC x BT = (0,a,0) x (-a/2, a/2, a/√2) = (a(a/√2), 0, a(-a/2)) = (a^2/√2, 0, -a^2/2).
Again, the same normal vector. There must be a misunderstanding in calculating the normal vector for TBC or the definition of the planes.

Let's use a geometric approach again. Let M be the midpoint of AD and N be the midpoint of BC. TM is perpendicular to AD, TN is perpendicular to BC. TM = TN = a√3/2. MN = a.
Consider the plane passing through T, M, and N. This plane is perpendicular to AD and BC.
The line of intersection of planes TAD and TBC is a line through T parallel to AD and BC.
Let's take a plane that cuts through the pyramid perpendicular to the line of intersection of TAD and TBC.

A simpler approach for regular tetrahedron or pyramid with equilateral triangle faces:
Consider the projection of T onto the base, which is O. TO = a/√2.
Consider the midpoint of AD, M. AM = a/2. TM = a√3/2.
In triangle TOM, TM^2 = TO^2 + OM^2. (a√3/2)^2 = (a/√2)^2 + OM^2.
3a^2/4 = a^2/2 + OM^2. OM^2 = 3a^2/4 - 2a^2/4 = a^2/4. OM = a/2. This is correct as O is the center of the square.

The angle between the planes TAD and TBC is the angle between their normal vectors. Or, by definition of dihedral angle, it is the angle between two lines, one in each plane, that are perpendicular to the line of intersection of the planes.

The line of intersection of TAD and TBC is parallel to AD and BC. Let's consider a plane perpendicular to AD and BC that passes through T. This plane contains the altitude TO and the midpoints of AD (M) and BC (N). So, the plane is TMN.

The angle between the planes TAD and TBC is the angle between TM and TN. However, TM and TN are not necessarily perpendicular to the line of intersection.

Let's reconsider the normal vectors.
Plane TAD: A=(0,0,0), D=(0,a,0), T=(a/2, a/2, a/√2). Vector AD = (0,a,0), Vector DT = (a/2, -a/2, a/√2).
Normal n1 = AD x DT = (0,a,0) x (a/2, -a/2, a/√2) = (a^2/√2, 0, -a^2/2). Simplify to (√2, 0, -1).

Plane TBC: B=(a,0,0), C=(a,a,0), T=(a/2, a/2, a/√2). Vector BC = (0,a,0), Vector CT = (a/2-a, a/2-a, a/√2) = (-a/2, -a/2, a/√2).
Normal n2 = BC x CT = (0,a,0) x (-a/2, -a/2, a/√2) = (a^2/√2, 0, a^2/2). Simplify to (√2, 0, 1).

Now we have two normal vectors: n1 = (√2, 0, -1) and n2 = (√2, 0, 1).
The cosine of the angle theta between the planes is given by the absolute value of the cosine of the angle between their normal vectors:
cos(theta) = |n1 . n2| / (||n1|| ||n2||).
n1 . n2 = (√2)(√2) + (0)(0) + (-1)(1) = 2 + 0 - 1 = 1.
||n1|| = sqrt((√2)^2 + 0^2 + (-1)^2) = sqrt(2 + 0 + 1) = sqrt(3).
||n2|| = sqrt((√2)^2 + 0^2 + 1^2) = sqrt(2 + 0 + 1) = sqrt(3).

cos(theta) = |1| / (sqrt(3) * sqrt(3)) = 1 / 3.

So, the cosine of the angle between the planes TAD and TBC is 1/3.