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 is expressed as the function: A(Θ) = 16 sin Θ • (cos Θ + 1). Assuming that the angle Θ = 45°, what is the area of the opening?

Answer 1

Answer:

A(45°) = 8*(1 + √2)  = 19.314 in^2

Step-by-step explanation:

Given:-

- The area A of the opening is given as:

                         A(Θ) = 16 sin Θ • (cos Θ + 1)

Find:-

Assuming that the angle Θ = 45°, what is the area of the opening?

Solution:-

- We can use the expression given for Area A opening and substitute the value of bent-up angle of (Θ = 45°) and compute the value of A(Θ). So we have:

                        A(Θ) = 16*sin (Θ) • (cos (Θ) + 1)

- Plug in the value:

                        A(45°) = 16*sin (45°) • (cos (45°) + 1)

                        A(45°) = 16*( 1 / √2 ) • ( 1 / √2 + 1)

                        A(45°) = 16*( 1 / √2 ) • ( (1 + √2)  / √2 )

                        A(45°) = 16*( (1 + √2)  / 2 )

                        A(45°) = 8*(1 + √2)  = 19.314 in^2

Answer 2

We are given a rain gutter made from folding a 12 inch wide aluminum sheet to an angle theta. The area of the opening is expressed as  A(Θ) = 16 sin Θ (cos Θ + 1) where Θ = 45°. we can just substitute directly the angle in degrees mode in the calculator and the answer is equal to 19.32 inches2.