1. **State the problem:** We have a right triangle with sides 4 (vertical), 4\sqrt{3} (horizontal), and 8 (hypotenuse), and angles 30° and 60°. We want to find $\sin 60^\circ$ and $\cos 60^\circ$ using this triangle. 2. **Recall definitions:** - $\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}$ - $\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}$ 3. **Identify sides for 60° angle:** - Opposite side to 60° is the vertical side = 4 - Adjacent side to 60° is the horizontal side = 4\sqrt{3} - Hypotenuse = 8 4. **Calculate $\sin 60^\circ$:** $$\sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{8}$$ Simplify: $$\sin 60^\circ = \frac{\cancel{4}}{\cancel{8}} = \frac{1}{2}$$ 5. **Calculate $\cos 60^\circ$:** $$\cos 60^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4\sqrt{3}}{8}$$ Simplify: $$\cos 60^\circ = \frac{\cancel{4}\sqrt{3}}{\cancel{8}} = \frac{\sqrt{3}}{2}$$ 6. **Final answers:** - $\sin 60^\circ = \frac{1}{2}$ - $\cos 60^\circ = \frac{\sqrt{3}}{2}$