Question

a real - world problem
you are told that in a billiards shot, the cue ball was shot at the eight ball, which was 8 inches away. as a result, the eight ball rolled into a pocket, which was 6 inches away.
knowing that the angle made with the path of the cue ball and the resulting path of the eight ball is larger than 90°, it can be determined that the original distance from the cue ball to the pocket was greater than
inches.

Explanation:

Step1: Apply the Law of Cosines

Let \(a = 8\), \(b = 6\), \(c\) be the distance from the cue - ball to the pocket, and \(\theta\) be the angle between the path of the cue - ball and the path of the eight - ball. By the Law of Cosines, \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\).

Step2: Analyze the cosine of the angle

Since \(\theta>90^{\circ}\), then \(\cos\theta<0\). Substituting \(a = 8\) and \(b = 6\) into the Law of Cosines formula: \(c^{2}=8^{2}+6^{2}-2\times8\times6\times\cos\theta=64 + 36-96\cos\theta=100-96\cos\theta\).

Step3: Determine the range of \(c\)

Because \(\cos\theta<0\), then \(-96\cos\theta>0\). So \(c^{2}=100-96\cos\theta>100\). Taking the square root of both sides (and considering \(c>0\) since \(c\) represents a distance), we get \(c > 10\).

Answer:

\(10\)