A rain gutter is to be constructed of aluminum sheets 12 inches wide. After marking off a length of 4 inches from each edge, this length is bent up at an angle θ. The area A of the opening may be expressed as the function: A(θ) = 16 sin θ ⋅ (cos θ + 1). If θ = 45°, what is the area of the opening?

Accepted Solution

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]