Answer:
[tex]4(2+\sqrt{2})\text{ square unit}[/tex]
Step-by-step explanation:
Given function that shows the area of the opening,
[tex]A(\theta)=16 \sin\theta (\sin \theta + 1)[/tex]
If [tex]\theta = 45^{\circ}[/tex]
Hence, the area of the opening would be,
[tex]A(45^{\circ})=16 \sin 45^{\circ} (\cos 45^{\circ} + 1)[/tex]
[tex]=16\times \frac{1}{\sqrt{2}}\times (\frac{1}{\sqrt{2}}+1)[/tex]
[tex]=16(\frac{1}{2}+\frac{1}{\sqrt{2}})[/tex]
[tex]=8+\frac{16}{\sqrt{2}}[/tex]
[tex]=8+4\sqrt{2}[/tex]
[tex]=4(2+\sqrt{2})\text{ square unit}[/tex]
