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

For this problem, I think there is no need for the details of 12 inches width and 4 inches length. This is because an equation is already given. It was clearly specified that A as a function of θ represents the area of the opening. Then, we are asked to find exactly that: the area of opening. Moreover, the value of θ was also given. Therefore, I am quite sure that the initial details given are for the purpose of red herring only.

So, all we have to do is substitute θ=45° to the function given. 
A = 16 sin 45° ⋅ (cos 45° + 1)

The angle 45° is a special angle in trigonometry. So, it would be easy to remember trigonometric functions of this angle. Sine of 45° is equal to √2/2 while cosine of 45° is also √2/2.

A = 16(√2/2) ⋅ (√2/2 + 1)
A = 8+8√2
A = 19.31 square inches

Answer 2

Refer to the diagram shown below which illustrates the problem.

From simple geometry, obtain
a = 4 cosθ
b = 4 sin θ

Therefore the area of the opening is
A = (1/2)*(2a + 4 + 4)*(b)
   = (a + 4)*b
   = (4a cosθ + 4)*(4a sin θ)
   = 16(1+ cosθ)sin θ
This agrees with the given area.

When θ = 45°, sinθ = 1/√2, cos θ = 1/√2.
Therefore
A = 16(1 + 1/√2)*(1/√2)
    = 16(1/√2 + 1/2) = 16(√2/2 + 1/2)
    = 8(√2 + 1) = 19.3 in²

Answer: 8(1 + √2) in², or 19.3 in²