1. **State the problem:** We need to find the cotangent of angle $\angle S$ in right triangle $RQS$ where $\angle R$ is the right angle. 2. **Given sides:** - $RQ = 3\sqrt{14}$ (horizontal side) - $SQ = 2\sqrt{42}$ (hypotenuse) - $RS$ is the vertical side (unknown length) 3. **Find the missing side $RS$ using the Pythagorean theorem:** $$SQ^2 = RS^2 + RQ^2$$ $$\Rightarrow RS^2 = SQ^2 - RQ^2$$ Calculate each term: $$SQ^2 = (2\sqrt{42})^2 = 4 \times 42 = 168$$ $$RQ^2 = (3\sqrt{14})^2 = 9 \times 14 = 126$$ So, $$RS^2 = 168 - 126 = 42$$ $$RS = \sqrt{42}$$ 4. **Recall the definition of cotangent:** $$\cot(\theta) = \frac{\text{adjacent side}}{\text{opposite side}}$$ For $\angle S$, the adjacent side is $RS$ and the opposite side is $RQ$. 5. **Calculate $\cot(\angle S)$:** $$\cot(S) = \frac{RS}{RQ} = \frac{\sqrt{42}}{3\sqrt{14}}$$ 6. **Simplify the expression:** First, simplify the denominator: $$3\sqrt{14} = 3 \times \sqrt{14}$$ Rewrite $\sqrt{42}$ and $\sqrt{14}$: $$\sqrt{42} = \sqrt{6 \times 7}$$ $$\sqrt{14} = \sqrt{7 \times 2}$$ Divide inside the fraction: $$\frac{\sqrt{42}}{3\sqrt{14}} = \frac{\sqrt{42}}{3\sqrt{14}} = \frac{\sqrt{42}}{3\sqrt{14}}$$ Divide numerator and denominator inside the square roots: $$= \frac{\sqrt{42}}{3\sqrt{14}} = \frac{\sqrt{42}}{3\sqrt{14}} = \frac{\sqrt{42}}{3\sqrt{14}}$$ Alternatively, rationalize the denominator: $$\cot(S) = \frac{\sqrt{42}}{3\sqrt{14}} \times \frac{\sqrt{14}}{\sqrt{14}} = \frac{\sqrt{42} \times \sqrt{14}}{3 \times 14} = \frac{\sqrt{588}}{42}$$ Simplify $\sqrt{588}$: $$588 = 49 \times 12$$ $$\sqrt{588} = \sqrt{49 \times 12} = 7\sqrt{12} = 7 \times 2\sqrt{3} = 14\sqrt{3}$$ So, $$\cot(S) = \frac{14\sqrt{3}}{42} = \frac{\cancel{14}\sqrt{3}}{\cancel{42}3} = \frac{\sqrt{3}}{3}$$ 7. **Final answer:** $$\boxed{\cot(S) = \frac{\sqrt{3}}{3}}$$