The coordinates of a point is the location of the point in a plane.
The coordinates of the centers of holes are: (48.5, 28) and (28, 48.5)
Given
[tex]\theta_1 = \theta_2 = \theta_3 = 30^o[/tex]
[tex]R = 60[/tex]
[tex]r = 56[/tex]
I've added an attachment as an illustration
Considering [tex](x_1,y_1)[/tex]
To solve for x1, we make use of cosine ratio.
So, we have:
[tex]\cos(\theta_1) =\frac{x}{r}[/tex]
Make x the subject
[tex]x_1 = r \times \cos(\theta_1)[/tex]
[tex]x_1 = 56 \times \cos(30^o)[/tex]
[tex]x_1 = 48.5[/tex]
To solve for y1, we make use of sine ratio.
So, we have:
[tex]\sin(\theta_1) =\frac{y_1}{r}[/tex]
Make y the subject
[tex]y_1 = r \times \sin(\theta_1)[/tex]
[tex]y_1 = 56 \times \sin(30^o)[/tex]
[tex]y_1 = 28[/tex]
So, we have:
[tex](x_1,y_1) = (48.5,28)[/tex]
Considering [tex](x_2,y_2)[/tex]
To solve for x2, we make use of cosine ratio.
So, we have:
[tex]\cos(\theta_1+\theta_2) =\frac{x_2}{r}[/tex]
Make x the subject
[tex]x_2 = r \times \cos(\theta_1+\theta_2)[/tex]
[tex]x_2 = 56 \times \cos(30+30)[/tex]
[tex]x_2 = 56 \times \cos(60^o)[/tex]
[tex]x_2 = 28[/tex]
To solve for y1, we make use of sine ratio.
So, we have:
[tex]\sin(\theta_1+\theta_2) =\frac{y_2}{r}[/tex]
Make y the subject
[tex]y_2 = r \times \sin(\theta_1+\theta_2)[/tex]
[tex]y_2 = 56 \times \sin(30+30)[/tex]
[tex]y_2 = 56 \times \sin(60)[/tex]
[tex]y_2 = 48.5[/tex]
So, we have:
[tex](x_2,y_2) = (28,48.5)[/tex]
Hence, the coordinates of the centers of the holes are: (48.5, 28) and (28, 48.5)
Read more about coordinate geometry at:
https://brainly.com/question/8121530
