1. **Problem statement:** We need to find the length of the hypotenuse in a right-angled triangle where one angle is 60° and the side opposite this angle is 6 meters. 2. **Formula used:** In a right triangle, the sine of an angle is the ratio of the length of the side opposite the angle to the hypotenuse. This is given by: $$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$$ 3. **Apply the formula:** Here, $\theta = 60^\circ$, opposite side = 6 m, and hypotenuse = $h$ (unknown). So, $$\sin(60^\circ) = \frac{6}{h}$$ 4. **Recall the value of $\sin(60^\circ)$:** $$\sin(60^\circ) = \frac{\sqrt{3}}{2}$$ 5. **Set up the equation:** $$\frac{\sqrt{3}}{2} = \frac{6}{h}$$ 6. **Solve for $h$:** Multiply both sides by $h$ and then divide both sides by $\frac{\sqrt{3}}{2}$: $$h = \frac{6}{\frac{\sqrt{3}}{2}} = 6 \times \frac{2}{\sqrt{3}} = \frac{12}{\sqrt{3}}$$ 7. **Simplify the expression:** Rationalize the denominator: $$h = \frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}$$ **Final answer:** The length of the hypotenuse is $4\sqrt{3}$ meters.