anonymous
  • anonymous
SIMPLE TRIG PROBLEMS NEED HELP. I WILL GIVE MEDAL AND FAN
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
@Nnesha
anonymous
  • anonymous
Pls help

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anonymous
  • anonymous
@Preetha
anonymous
  • anonymous
@pooja195
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
Not really sure what the first one is asking for, but I can help you with the second one.
anonymous
  • anonymous
That sound good?
anonymous
  • anonymous
Well I mean, it's really simple. The angle \(\theta\) can be found using the formula \[\large \sf \theta~=~\frac{arc~length}{radius}\]
anonymous
  • anonymous
Here's a nice little cheat-sheet for trig formulas: http://learnix.net/wordpress/wp-content/uploads/Trig-Cheat-Sheet-1.4.pdf
anonymous
  • anonymous
ok
anonymous
  • anonymous
i do not have arc lenght
anonymous
  • anonymous
Ugh, this is what I get for trying to do math while sleep deprived. Give me a minute, sorry.
anonymous
  • anonymous
lol its ok
mathstudent55
  • mathstudent55
Let's look at the first problem. You are given an angular velocity, and you asked to find a linear velocity.
anonymous
  • anonymous
yes
anonymous
  • anonymous
are you still there
mathstudent55
  • mathstudent55
In the formula below, we use \(\omega\) (small-case Greek letter "omega" is usually used to denote an angular velocity). Point A moves in a circular path around point O with an angular velocity of \(\omega\). If point A is r units from point O, then the linear velocity of point A is \(v = r \omega \). \(\omega\) must be expressed in radians per unit time in this formula.
mathstudent55
  • mathstudent55
\(\ \large v = r\omega\) In your case, you have the angular velocity given in revolutions per minute. You need to convert revolutions into radians.
mathstudent55
  • mathstudent55
Since the answer must be in miles per hour, you will also need to convert minutes into hours and inches into miles.
anonymous
  • anonymous
so angular velocity is 6500
anonymous
  • anonymous
i dont get it, what do i do next
anonymous
  • anonymous
@mathstudent55
mathstudent55
  • mathstudent55
We start with this: \(\large v = r \omega = 7.25 ~in. \times ~3250 ~\dfrac{rev}{min}\)
mathstudent55
  • mathstudent55
Now we need to apply all the conversion factors.
anonymous
  • anonymous
but isnt angular velocity 6500
mathstudent55
  • mathstudent55
Angular velocity needs to be in radians per minute. You have revolutions per minute. 1 revolution is \(2 \pi\) radians, not 2 radians.
anonymous
  • anonymous
ohhhhh
mathstudent55
  • mathstudent55
Let's deal with the conversion of revolutions to radians. It's in red below. \(\large v = r \omega = 7.25 ~in. \times ~3250 ~\dfrac{rev}{min} \color{red}{\times \dfrac{2 \pi ~rad }{1~rev}}\)
anonymous
  • anonymous
Since you guys are still working on the first part, I'll just leave this here so you can work it out after you finish. You can find arc length from sector area and radius with for formula \[\large \sf Area~=~\frac{1}{2} \times r \times Arc \huge \rightarrow \large \frac{2 \times Area}{r}~=~Arc\] from there, you can find \(\large \sf \theta\) with the original formula. \[\large \sf \theta~=~\frac{Arc~length}{Radius}\]
mathstudent55
  • mathstudent55
Now we need to deal with the conversion of inches into miles. We can convert using 12 in. = 1 ft, and 5280 ft = 1 mile. This conversion is in green below. \(v = r \omega = 7.25 ~in. \color{green}{\times \dfrac{1~ft}{12~in.} \times \dfrac{1~mile}{5280~ft} } \times ~3250 ~\dfrac{rev}{min} \color{red}{\times \dfrac{2 \pi ~rad }{1~rev}}\) Now we have miles per minute. We need to convert minutes to hours.
mathstudent55
  • mathstudent55
The conversion from minutes to hours is in blue below. 60 minutes = 1 hour \(v = r \omega = 7.25 ~in. \color{green}{\times \dfrac{1~ft}{12~in.} \times \dfrac{1~mile}{5280~ft} } \times ~3250 ~\dfrac{rev}{min} \color{red}{\times \dfrac{2 \pi ~rad }{1~rev}} \color{blue} {\times \dfrac{60~min}{1~hour} }\)
mathstudent55
  • mathstudent55
What do you get? \(v = \dfrac{7.25 \times 3250 \times 2 \times \pi \times 60}{12 \times 5280} \dfrac{miles}{hour} \)
anonymous
  • anonymous
one sec
anonymous
  • anonymous
140.2
mathstudent55
  • mathstudent55
That's what I got.
anonymous
  • anonymous
140.2 miles per hour
anonymous
  • anonymous
OK @LegendarySadist
anonymous
  • anonymous
so whats my first step
anonymous
  • anonymous
That would be to find the arc length \[\large \sf Area~=~\frac{1}{2} \times r \times Arc \huge \rightarrow \large \frac{2 \times Area}{r}~=~Arc\]
anonymous
  • anonymous
so arc length is 20/3
anonymous
  • anonymous
Correct. Now we can plug that into our other formula to find \(\large \sf \theta\) \[\large \sf \theta~=~\frac{Arc~length}{Radius}\]
anonymous
  • anonymous
bit i need theta in radians
anonymous
  • anonymous
but*
anonymous
  • anonymous
This is the solution in radians. That's why I chose this formula instead of the degrees formula.
anonymous
  • anonymous
i got 20/27
anonymous
  • anonymous
That would be it.
mathstudent55
  • mathstudent55
Second problem: \(\Large A_{sector} = \dfrac{\theta}{2 \pi ~rad}\pi r^2\) \(\Large \theta = \dfrac{2 \pi A}{\pi r^2} = \dfrac{2A}{r^2} = \dfrac{2 \times 30 ft^2}{(9~ft)^2} = \dfrac{60}{81} = \dfrac{20}{27}\)

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