solving a real-world problem real-world connection you are told that in…

問題

solving a real-world problem
real-world connection
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 $\boldsymbol{square}$ inches.
cue ball
eight ball
pocket
8 in.
6 in.

解法ステップ

Understand the question
solving a real-world problem
real-world connection
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 $\boldsymbol{square}$ inches.
cue ball
eight ball
pocket
8 in.
6 in.
Explanation
Step1: Identify triangle sides/angle

We have a triangle with sides $a=8$ in, $b=6$ in, included angle $\theta=90^\circ+90^\circ=180^\circ$? No, correct: the angle between cue ball path and 8-ball path is >90°, use Law of Cosines. Let original distance be $c$, angle $\alpha>90^\circ$.
Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos\alpha$

Step2: Analyze $\cos\alpha$ for $\alpha>90^\circ$

For $\alpha>90^\circ$, $\cos\alpha < 0$, so $-2ab\cos\alpha > 0$.

Step3: Compare to right triangle case

If $\alpha=90^\circ$, $c^2=8^2+6^2=64+36=100$, $c=10$ in.
Since $-2ab\cos\alpha>0$, $c^2 = 100 + |2ab\cos\alpha| > 100$, so $c>10$.
Final answer
10

答え

Explanation

Step1: Identify triangle sides/angle

We have a triangle with sides $a=8$ in, $b=6$ in, included angle $\theta=90^\circ+90^\circ=180^\circ$? No, correct: the angle between cue ball path and 8-ball path is >90°, use Law of Cosines. Let original distance be $c$, angle $\alpha>90^\circ$.
Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos\alpha$

Step2: Analyze $\cos\alpha$ for $\alpha>90^\circ$

For $\alpha>90^\circ$, $\cos\alpha < 0$, so $-2ab\cos\alpha > 0$.

Step3: Compare to right triangle case

If $\alpha=90^\circ$, $c^2=8^2+6^2=64+36=100$, $c=10$ in.
Since $-2ab\cos\alpha>0$, $c^2 = 100 + |2ab\cos\alpha| > 100$, so $c>10$.

Answer

10

Question Image

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Question Analysis

Subject mathematics

Sub Subject geometry

Education Level high school

Difficulty unspecified

Question Type with image, calculation

Multi Question No

Question Count 1

Analysis Status completed

Analyzed At 2026-02-09T20:15:02

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solving a real-world problem
real-world connection
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 $\boldsymbol{square}$ inches.
cue ball
eight ball
pocket
8 in.
6 in.

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