anonymous
  • anonymous
@rvc
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I need your help :/
anonymous
  • anonymous
i need medals
rvc
  • rvc
yes @nopen

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More answers

anonymous
  • anonymous
it's part c
anonymous
  • anonymous
I got the answer for a and b
anonymous
  • anonymous
Part a.) is OC = OT - CT = R- r
anonymous
  • anonymous
Part b.) is R*sin*theta = r(1+ sin theta)
rvc
  • rvc
im extremely sorry gtg now
Michele_Laino
  • Michele_Laino
we can write this: \[\Large \begin{gathered} 21 = 2R + R2\theta = 2R + 2R\arcsin \left( {\frac{3}{4}} \right) = \hfill \\ \hfill \\ = 2R\left\{ {1 + \arcsin \left( {\frac{3}{4}} \right)} \right\} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
since: \[\Large ATB = R2\theta \]
anonymous
  • anonymous
yea I did that but my problem is why am I to taking the 1 with the given angle?
Michele_Laino
  • Michele_Laino
by definition of radians, we have: \[\Large L = \alpha R\] |dw:1436029827922:dw|
dan815
  • dan815
http://prntscr.com/7orri3
anonymous
  • anonymous
@dan815 your answer is wrong =_=
anonymous
  • anonymous
@Michele_Laino I do remember that arc length formula but what I'm asking is that, if we are given that O is sin theta = 3/4 why are we taking the 1?
dan815
  • dan815
what is the answer
anonymous
  • anonymous
2.43
anonymous
  • anonymous
@Michele_Laino what you've written in your answer is exactly the same thing that is written in the solution bank I simply wanna know why you talking (1 + arcsin(3/4) instead of just arcsin3/4?
anonymous
  • anonymous
*are taking
Michele_Laino
  • Michele_Laino
since I have factored out the quantity 2R
dan815
  • dan815
hey michele can you tell me what's wrong in my method, i cant find it
Michele_Laino
  • Michele_Laino
your link tells me "500 Internal Server Error" @dan815
anonymous
  • anonymous
lol
dan815
  • dan815
1 Attachment
dan815
  • dan815
how about that one
anonymous
  • anonymous
Ohhhhh Ok I got it!!
dan815
  • dan815
oh oops its the whole perimeter not just the arc perimeter
anonymous
  • anonymous
THANK YOU @Michele_Laino! :DDDDD
Michele_Laino
  • Michele_Laino
yes! the perimeter is the perimeter of the sector @dan815
Michele_Laino
  • Michele_Laino
:) @nopen
dan815
  • dan815
http://prntscr.com/7orzms hehe too clumpsy

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