Generated by claude-sonnet-4-6 · 2026-06-15 10:21 · grounding rag
Model Answer
Answer: (B) $2\sqrt{2}$
Given $\sin A = \dfrac{1}{3}$, so opposite = $k$, hypotenuse = $3k$. By Pythagoras, adjacent $= \sqrt{9k^2 - k^2} = 2\sqrt{2}\,k$.
$$\cot A = \frac{\text{adjacent}}{\text{opposite}} = \frac{2\sqrt{2}\,k}{k} = 2\sqrt{2}$$
Source: Chapter 8, Section 8.2
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Explanation
- The textbook itself uses $\sin A = \frac{1}{3}$ as the worked example in Section 8.2, finding $AB = 2\sqrt{2}\,k$.
- $\cot A = \frac{\cos A}{\sin A} = \frac{\text{adjacent}}{\text{opposite}}$, not $\frac{\text{adjacent}}{\text{hypotenuse}}$ — a common mix-up to avoid.
- Option A ($\frac{2\sqrt{2}}{3}$) is $\cos A$, not $\cot A$ — watch which ratio is asked.