Suppose triangle ABC has vertices at A(1, 0), B(10, 0), and C(2, 6). After a 60° counterclockwise rotation about the origin, vertex B' has coordinates (5, ?).
recall that
x' = x cosθ - y sinθ
y' = x sinθ + y cosθ
To find the y-coordinate of vertex B' after a 60° counterclockwise rotation about the origin, we need to apply the rotation matrix.
The rotation matrix for a counterclockwise rotation of θ degrees about the origin is:
[ cos(θ) -sin(θ) ]
[ sin(θ) cos(θ) ]
In this case, since we're rotating 60° counterclockwise, our rotation matrix is:
[ cos(60°) -sin(60°) ]
[ sin(60°) cos(60°) ]
To find the coordinates of B' after the rotation, we need to multiply the rotation matrix with the coordinates of B (10, 0):
[ cos(60°) -sin(60°) ] * [ 10 ]
[ sin(60°) cos(60°) ] [ 0 ]
Calculating the matrix multiplication:
[ cos(60°) -sin(60°) ] * [ 10 ] = [ cos(60°)*10 + (-sin(60°))*0 ]
[ sin(60°) cos(60°) ] [ 0 ] [ sin(60°)*10 + cos(60°)*0 ]
Simplifying:
[ cos(60°) -sin(60°) ] * [ 10 ] = [ cos(60°)*10 + 0 ]
[ sin(60°) cos(60°) ] [ 0 ] [ sin(60°)*10 + 0 ]
[ cos(60°) -sin(60°) ] * [ 10 ] = [ (1/2)*10 + 0 ]
[ sin(60°) cos(60°) ] [ 0 ] [ sqrt(3)/2*10 + 0 ]
Simplifying further:
[ cos(60°) -sin(60°) ] * [ 10 ] = [ 5 ]
[ sin(60°) cos(60°) ] [ 0 ] [ 5*sqrt(3) ]
Therefore, the coordinates of B' after the rotation are (5, 5*sqrt(3)).