Question

70. Machine Shop Calculations A steel plate has the

Answer 1

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

$\theta_1 = \theta_2 = \theta_3 = 30^o$

$R = 60$

$r = 56$

I've added an attachment as an illustration

Considering $(x_1,y_1)$

To solve for x1, we make use of cosine ratio.

So, we have:

$\cos(\theta_1) =\frac{x}{r}$

Make x the subject

$x_1 = r \times \cos(\theta_1)$

$x_1 = 56 \times \cos(30^o)$

$x_1 = 48.5$

To solve for y1, we make use of sine ratio.

So, we have:

$\sin(\theta_1) =\frac{y_1}{r}$

Make y the subject

$y_1 = r \times \sin(\theta_1)$

$y_1 = 56 \times \sin(30^o)$

$y_1 = 28$

So, we have:

$(x_1,y_1) = (48.5,28)$

Considering $(x_2,y_2)$

To solve for x2, we make use of cosine ratio.

So, we have:

$\cos(\theta_1+\theta_2) =\frac{x_2}{r}$

Make x the subject

$x_2 = r \times \cos(\theta_1+\theta_2)$

$x_2 = 56 \times \cos(30+30)$

$x_2 = 56 \times \cos(60^o)$

$x_2 = 28$

To solve for y1, we make use of sine ratio.

So, we have:

$\sin(\theta_1+\theta_2) =\frac{y_2}{r}$

Make y the subject

$y_2 = r \times \sin(\theta_1+\theta_2)$

$y_2 = 56 \times \sin(30+30)$

$y_2 = 56 \times \sin(60)$

$y_2 = 48.5$

So, we have:

$(x_2,y_2) = (28,48.5)$

Hence, the coordinates of the centers of the holes are: (48.5, 28) and (28, 48.5)

Read more about coordinate geometry at:

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