# In a soccer game, Wayne is lining up for a free kick. The goal posts of the net are 12 m apart. The ball is 25.8 m from one post and 22.5 m from the other post. Sketch a diagram to represent the position of the ball and the two goal posts. Label the triangle with the given measurements. Calculate the angle formed by the ball and the two goal posts to the nearest whole degree.

We are given that the goalposts are 12 m apart and that the ball is 25.8 m from one post and 22.5 m from the other post.

(See attachment for sketch.)

Let the goalposts be marked A and B and the ball at E. Let AC=25.8 m, and let BC=22.5 m. Then, using normal assignments, we have a=22.5, b=25.8, e=12, and we are looking for the measure of angle E.

Since we have three sides of the triangle, we use the Law of Cosines.

Then, `e^2=a^2+b^2-2abcosE`, or `cosE=(e^2-a^2-b^2)/(-2ab)`.

Here, we get `12^2=25.8^2+22.5^2-2(25.8)(22.5)cosE`

or `cosE~~.8853`

Then, `E=cos^(-1)(.8853)~~27.7^(@)`.

(We use the inverse cosine or arccos to find the angle using a scientific calculator or computer application.)

Thus, the angle we seek is about `28^(@)` .

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Now that we have the angle formed by the ball and the posts (angle E), we could find the measures of angle A and B in a similar fashion. But we could also use the Law of Sines:

`a/(sinA)=b/(sinB)=e/(sinE)`.

We could not use the Law of Sines earlier as we did not have any of the angles, so every proportion we set up would have 2 unknowns.

Using care with rounding, we see that E is just smaller than 28 degrees, A is close to 62 degrees, and B is close to 90 degrees (just a little more than 90.) This matches with the largest side being opposite the largest angle in a triangle.

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