1. **State the problem:** Given $x=3\tan\theta$, we want to label all sides of a right triangle and find the sides in terms of $x$ and $\theta$. 2. **Recall the definition of tangent:** In a right triangle, $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$. 3. **Assign sides:** Let the opposite side be $O$, the adjacent side be $A$, and the hypotenuse be $H$. 4. **Express $x$ in terms of sides:** Since $x=3\tan\theta$, we have $x = 3 \frac{O}{A}$. 5. **Solve for $O$:** Multiply both sides by $A$: $$xA = 3O$$ Cancel common factor 3: $$\cancel{3}O = \frac{xA}{\cancel{3}}$$ So, $$O = \frac{xA}{3}$$ 6. **Use Pythagoras theorem:** $$H = \sqrt{O^2 + A^2} = \sqrt{\left(\frac{xA}{3}\right)^2 + A^2} = \sqrt{\frac{x^2 A^2}{9} + A^2} = \sqrt{A^2 \left(\frac{x^2}{9} + 1\right)} = A \sqrt{\frac{x^2}{9} + 1}$$ 7. **Summary:** - Opposite side $O = \frac{xA}{3}$ - Adjacent side $A$ (can be any positive length) - Hypotenuse $H = A \sqrt{\frac{x^2}{9} + 1}$ This labels all sides of the right triangle in terms of $x$ and $A$. **Final answer:** $$O = \frac{xA}{3}, \quad A = A, \quad H = A \sqrt{\frac{x^2}{9} + 1}$$