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: Identify triangle properties

We have a triangle with two sides: $a=8$ in, $b=6$ in, and the included angle $\theta > 90^\circ$. We need to find the minimum length of the third side $c$ (distance from cue ball to pocket).

Step2: Use Law of Cosines

The Law of Cosines states $c^2 = a^2 + b^2 - 2ab\cos\theta$.

Step3: Analyze $\cos\theta$ for $\theta>90^\circ$

For $\theta > 90^\circ$, $\cos\theta < 0$, so $-2ab\cos\theta > 0$. The minimum value of $c$ occurs when $\cos\theta$ is closest to 0 (i.e., $\theta=90^\circ$, where $\cos\theta=0$).

Step4: Calculate minimum $c$ at $\theta=90^\circ$

$$ c^2 = 8^2 + 6^2 - 2(8)(6)(0) = 64 + 36 = 100 $$

$$ c = \sqrt{100} = 10 $$

Since $\theta>90^\circ$, $c > 10$.

Answer:

10