1. **Problem 12:** Given a right triangle with a 60° angle, side adjacent to 60° is $\frac{5\sqrt{3}}{3}$, side opposite 60° is $y$, and hypotenuse is $x$. Find $x$ and $y$. 2. **Recall trigonometric ratios:** For angle $\theta=60^\circ$ in a right triangle, - Adjacent side = $\text{hypotenuse} \times \cos 60^\circ$ - Opposite side = $\text{hypotenuse} \times \sin 60^\circ$ 3. Use $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$. 4. Let hypotenuse be $x$, adjacent side is $\frac{5\sqrt{3}}{3}$: $$\cos 60^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5\sqrt{3}/3}{x} = \frac{1}{2}$$ 5. Solve for $x$: $$\frac{5\sqrt{3}}{3x} = \frac{1}{2} \implies 5\sqrt{3} \times 2 = 3x \implies 10\sqrt{3} = 3x$$ $$x = \frac{10\sqrt{3}}{3}$$ 6. Find $y$ using $\sin 60^\circ$: $$\sin 60^\circ = \frac{y}{x} = \frac{\sqrt{3}}{2}$$ $$y = x \times \frac{\sqrt{3}}{2} = \frac{10\sqrt{3}}{3} \times \frac{\sqrt{3}}{2} = \frac{10 \times 3}{3 \times 2} = \frac{10}{2} = 5$$ 7. So, $x = \frac{10\sqrt{3}}{3}$ and $y = 5$. --- 8. **Problem 14:** Right triangle with angle 30°, hypotenuse $16$, opposite side $m$, adjacent side $n$. Find $m$ and $n$. 9. Use trigonometric ratios for $30^\circ$: - $\sin 30^\circ = \frac{1}{2}$ - $\cos 30^\circ = \frac{\sqrt{3}}{2}$ 10. Opposite side $m = 16 \times \sin 30^\circ = 16 \times \frac{1}{2} = 8$ 11. Adjacent side $n = 16 \times \cos 30^\circ = 16 \times \frac{\sqrt{3}}{2} = 8\sqrt{3}$ **Final answers:** - Problem 12: $x = \frac{10\sqrt{3}}{3}$, $y = 5$ - Problem 14: $m = 8$, $n = 8\sqrt{3}$