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?
2 answers:
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
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
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