Question

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: Identify triangle properties

We have a triangle with two sides $a=8$ in, $b=6$ in, and the angle $\theta$ between them greater than $90^\circ$. Let the third side be $c$ (distance from cue ball to pocket).

Step2: Apply Law of Cosines

For any triangle, $c^2 = a^2 + b^2 - 2ab\cos\theta$. Since $\theta > 90^\circ$, $\cos\theta < 0$, so $-2ab\cos\theta > 0$.

Step3: Calculate threshold for right angle

When $\theta=90^\circ$, $\cos\theta=0$, so $c^2 = 8^2 + 6^2 = 64 + 36 = 100$, so $c=10$ in.

Step4: Analyze for obtuse angle

Since $-2ab\cos\theta > 0$, $c^2 > 8^2 + 6^2 = 100$, so $c > 10$.

Answer:

10