1. The problem involves finding the missing side lengths in right triangles with given angles and side lengths, specifically using 30° and 60° angles. 2. For a 30°-60°-90° triangle, the side ratios are important: the side opposite 30° is $x$, opposite 60° is $x\sqrt{3}$, and the hypotenuse is $2x$. 3. Using these ratios, we can find missing sides by setting the known side equal to one of these expressions and solving for $x$. 4. For problem 7, the adjacent side to 60° is 20, which corresponds to the side opposite 30°, so $x=20$. The opposite side to 60° is $x\sqrt{3} = 20\sqrt{3}$. 5. For problem 8, the vertical leg is $\frac{4\sqrt{3}}{3}$ opposite 60°. To find the hypotenuse, use $x\sqrt{3} = \frac{4\sqrt{3}}{3}$, so $x = \frac{4}{3}$. The hypotenuse is $2x = \frac{8}{3}$. 6. The handwritten multiplication $\frac{4\sqrt{3}}{3} \cdot \frac{\sqrt{3}}{3} = \frac{4 \times 3}{9} = \frac{4}{3}$ confirms the calculation. 7. For problem 9, the vertical leg opposite 60° is $\frac{7\sqrt{3}}{2}$. Using $x\sqrt{3} = \frac{7\sqrt{3}}{2}$, we get $x = \frac{7}{2}$. The hypotenuse is $2x = 7$. 8. For problem 10, the horizontal leg adjacent to 30° is 4, which corresponds to the side opposite 60° divided by $\sqrt{3}$. The handwritten multiplication $4 \cdot \frac{\sqrt{3}}{3} = \frac{4\sqrt{3}}{3}$ gives the side opposite 30°, so $X = \frac{4\sqrt{3}}{3}$. Final answers: - 7) Opposite side to 60°: $20\sqrt{3}$ - 8) Hypotenuse: $\frac{8}{3}$ - 9) Hypotenuse: $7$ - 10) $X = \frac{4\sqrt{3}}{3}$