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PLEASE HELP!! 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 = 90°, what is the area of the opening? When i plug 90 in i keep getting 15, but my choices are: 21.8 in2 16.0 in2 7.9 in2 1.0 in2

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@dangerousjesse any luck?

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Other answers:

  • phi
first, you can do this without a calculator. But if you do use a calculator, be sure it is in degree mode. you should(!) know sin(90º)= 1 and cos(90º)= 0
@phi i just figured it out actually!! C right? and could you please help me with one more!! im stuck on it :/
Simplify \[\sqrt{\left( 1-\sin \Theta \right)\left( 1+\sin \Theta \right)}\]
@phi i have no idea what to do
  • phi
First, How did you get C 7.9? It seems if sin(90) is 1 and cos(90) is 0 you would do 16 sin Θ • (cos Θ + 1) 16 • 1 •(0+1) and now we add 0+1 (not *too* hard) and multiply that by 16*1 you don't get 7.9
For some reason i divided 16 in 2 idk why haha thank you!!
  • phi
any idea what the answer is for your first question?
b right?
any idea on how to solve the second one? i really have no idea what to do @phi
  • phi
do you know how to multiply the two binomials inside the square root (some people use FOIL)
  • phi
if you don't see
yeah when i foil i get sqrt(1-sin O)
is that right? @phi
  • phi
can you show your work?
F: 1*1=1 O: 1*sinO=sinO I: -sinO*1=-sinO L: -sinO*sinO=-sin2 so its sqrt(1-sin^2) oops!
these are my options btw @phi ±sin Θ |cos Θ| ±tan Θ square root sine theta
  • phi
ok looks better there is a *very useful* identity you should remember for trig problems \[ \sin^2 x + \cos^2 x = 1 \] which you can rewrite as \[ \sin^2 x = 1 - \cos^2 x \] or \[ \cos^2 x= 1 - \sin^2 x \]
would it be B?
  • phi
yes. replace 1 -sin^2 with cos^2 inside the square root sqrt(cos^2) = cos is the only choice that looks close (they must only want the positive root) | cos Θ |
@phi thank you so so much!! i really truly appreciate it :)
  • phi

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