Answer: [tex]16\sqrt{3} \text{ in}^2[/tex]
Step-by-step explanation:
The length of the shorter side of the rectangle is [tex]8 \cos 60^{\circ}=4 \text{ in}[/tex].
The length of the longer side of the rectangle is [tex]8 \sin 60^{\circ}=4\sqrt{3} \text{ in}[/tex].
Therefore, the area of the rectangle is [tex]4 \cdot 4\sqrt{3}=16\sqrt{3} \text{ in}^2[/tex].
