A rain gutter is to be constructed of aluminum sheets 12 inches wide. After marking off a length of 4 inches from each edge, thi

Question

19 days ago

13

A rain gutter is to be constructed of aluminum sheets 12 inches wide. After marking off a length of 4 inches from each edge, thi

s 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?

Mathematics

2 answers:

zzz [12.3K] · Answer 1 · 2026-03-31T04:21:54+03:00

In this case, the dimensions of 12 inches in width and 4 inches in length are not crucial for solving the problem. An equation has been provided already, and it clearly states that A, as a function of θ, represents the area of the opening. We are specifically instructed to find this area, and the value of θ has been given. Therefore, the initial details seem to serve merely as a distraction.

Hence, we can directly substitute θ=45° into the given function.
A = 16 sin 45° ⋅ (cos 45° + 1)

Since 45° is a notable angle in trigonometry, remembering its functions is straightforward. The sine of 45° is √2/2, and the cosine of 45° is also √2/2.

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

AnnZ [12.3K] · Answer 2 · 2026-03-31T04:21:54+03:00

Look at the illustration below that depicts the situation.

From basic geometry, we find
a = 4 cos θ
b = 4 sin θ

Consequently, the area of the opening is
A = (1/2)*(2a + 4 + 4)*(b)
= (a + 4)*b
= (4a cos θ + 4)*(4a sin θ)
= 16(1+ cos θ)sin θ
This matches the provided area.

When θ = 45°, sin θ = 1/√2, cos θ = 1/√2.
Thus,
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