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.

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Aurora Harrell

Mathematics

12 January, 16:09

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.

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Simone Mccoy 12 January, 16:14
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Answer:10

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Lila Hinton · Answer 1

Assumptions: Based on your question is seems like you are trying to find the distance the cue ball was initially at from the pocket from the minimum angle being 90 degrees.

Answer: In this case we can assume the minimum angle that the eight ball had after the collision was 90 degrees and therefore we can use Pythagoras Theorem (only can be used for right triangles). This gives us the length of both sides of that triangle we have constructed which are 8 in. and 6 in ... In the context of our drawing we want to find the hypotenuse of the triangle which would be the initial distance the eight ball was from the pocket (once again being at the most minimum possible angle). a^2=b^2+c^2. a=sqrt (b^2+c^2).

Plug in for b and c and you should get that it was greater than 10 in.