The diagram in the Unit 1 Lesson 11, the description of the rotation of the image ∆P'O'G' is as follows;
1. C. Rotate 60° counterclockwise around O
The diagram of the triangle ∆POG consists of a triangle placed in a grid consisting of equilateral triangles.
The location of point O in both the pre-image and the image are (approximately) the same.
Therefore;
The point about which ∆POG is rotated is the point O.
The side OG in the pre-image is vertical, while O'G' in the image 1 is rotated to the right, through to the next adjacent leg of one of the triangles in the grid with vertex at O.
Given that the triangles in the grid are equilateral triangles, that have an 60° interior angles, we have;
The rotation transformation is a 60° counterclockwise rotation about the point O.
The correct option is therefore;
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