Question

solving 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 Pythagorean theorem for right - angled case

If the angle between the path of the cue ball to the eight - ball and the path of the eight - ball to the pocket was 90°, by the Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\), where \(a = 8\) inches and \(b = 6\) inches. So \(c=\sqrt{8^{2}+6^{2}}=\sqrt{64 + 36}=\sqrt{100}=10\) inches.

Step2: Consider the non - right angle case

Since the actual angle between the two paths is greater than 90°, in a non - right triangle, when the included angle \(\theta>90^{\circ}\), the length of the side opposite the non - right angle is longer than the hypotenuse of the right - angled triangle with the same two side lengths.

Answer:

10