Rotation Matrix

The image is under
the rotation of 60º counterclockwise
about the origin.

So write the rotation matrix
[cos 60º -sin 60º / sin 60º cos 60º].

Write the vertex matrix.

A(2, 2), B(4, 2), C(4, 4)

So the vertex matrix is
[2 4 4 / 2 2 4].

So the vertex matrix of the image is
[cos 60º -sin 60º / sin 60º cos 60º][2 4 4 / 2 2 4].

To find sin 60º and cos 60º,

draw a 30-60-90 triangle
whose sides are 1, √3, 2.

cos 60º

Cosine is CAH:
Cosine,
Adjacent side (1),
Hypotenuse (2).

So cos 60º = 1/2.

sin 60º

Sine is SOH:
Sine,
Opposite side (√3),
Hypotenuse (2).

So sin 60º = √3/2.

So -sin 60º = -√3/2.

sin 60º = √3/2

cos 60º = 1/2

Write the vertex matrix.

So
[cos 60º -sin 60º / sin 60º cos 60º][2 4 4 / 2 2 4]
= [1/2 -√3/2 / √3/2 1/2][2 4 4 / 2 2 4].

[1/2 -√3/2 / √3/2 1/2]
= (1/2)[1 -√3 / √3 1]

[2 4 4 / 2 2 4]
= 2[1 2 2 / 1 1 2]

Cancel the coefficients (1/2) and 2.

Solve [1 -√3 / √3 1][1 2 2 / 1 1 2].

Multiply Matrices

Row 1, column 1:
1⋅1 + (-√3)⋅1

Row 1, column 2:
1⋅2 + (-√3)⋅1

Row 1, column 3:
1⋅2 + (-√3)⋅2

Row 2, column 1:
√3⋅1 + 1⋅1

Row 2, column 2:
√3⋅2 + 1⋅1

Row 2, column 3:
√3⋅2 + 1⋅2

This is the vertex matrix of the image.

1⋅1 + (-√3)⋅1
= 1 - √3

1⋅2 + (-√3)⋅1
= 2 - √3

1⋅2 + (-√3)⋅2
= 2 - 2√3

√3⋅1 + 1⋅1
= √3 + 1

√3⋅2 + 1⋅1
= 2√3 + 1

√3⋅2 + 1⋅2
= 2√3⋅2 + 2

This is the vertex matrix of the image.

So column 1 is the image of A:
A'(1 - √3, 1 + √3).

Column 2 is the image of B:
B'(2 - √3, 1 + 2√3).

Column 3 is the image of C:
C'(2 - 2√3, 2 + 2√3).

So
A'(1 - √3, 1 + √3)
B'(2 - √3, 1 + 2√3)
C'(2 - 2√3, 2 + 2√3)
is the answer.

This is the graph of △ABC
and its image △A'B'C'.

The image is under
the rotation of 60º counterclockwise
about the origin.