Using the law of sines, it is found that the distance from the farthest soccer player to the goalie is 14.723 yards.
Suppose we have a triangle in which:
The lenghts and the sine of the angles are related as follows:
[tex]\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}[/tex]
In this problem, considering that the sum of the internal angles of a triangle is of 180º, the angle that the goalie makes with the players is given by:
[tex]A_g = 180 - (64 + 75) = 41[/tex]
Then, the situation can be represented by the graph given at the end of this problem.
Thus, applying the law of sines, we have that:
[tex]\frac{\sin{41^\circ}}{10} = \frac{\sin{75^\circ}}{d_{p1}} = \frac{\sin{64^\circ}}{d_{p2}}[/tex]
Then, the player 1 distance is:
[tex]\frac{\sin{41^\circ}}{10} = \frac{\sin{75^\circ}}{d_{p1}}[/tex]
[tex]d_{p1} = \frac{10\sin{75^\circ}}{\sin{41^\circ}}[/tex]
[tex]d_{p1} = 14.723[/tex]
The player 2 distance is:
[tex]\frac{\sin{41^\circ}}{10} = \frac{\sin{64^\circ}}{d_{p2}}[/tex]
[tex]d_{p2} = \frac{10\sin{75^\circ}}{\sin{64^\circ}}[/tex]
[tex]d_{p2} = 10.747[/tex]
The farthest distance is of 14.723 yards, as 14.723 > 10.747.
You can learn more about the law of sines at https://brainly.com/question/12827625
