Question

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?

Answer 1

Answer:

$4(2+\sqrt{2})\text{ square unit}$

Step-by-step explanation:

Given function that shows the area of the opening,

$A(\theta)=16 \sin\theta (\sin \theta + 1)$

If $\theta = 45^{\circ}$

Hence, the area of the opening would be,

$A(45^{\circ})=16 \sin 45^{\circ} (\cos 45^{\circ} + 1)$

$=16\times \frac{1}{\sqrt{2}}\times (\frac{1}{\sqrt{2}}+1)$

$=16(\frac{1}{2}+\frac{1}{\sqrt{2}})$

$=8+\frac{16}{\sqrt{2}}$

$=8+4\sqrt{2}$

$=4(2+\sqrt{2})\text{ square unit}$