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.

**Further Reading**