SIMPLE TRIG PROBLEMS NEED HELP. I WILL GIVE MEDAL AND FAN

- anonymous

SIMPLE TRIG PROBLEMS NEED HELP. I WILL GIVE MEDAL AND FAN

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- schrodinger

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- anonymous

@Nnesha

- anonymous

Pls help

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## More answers

- anonymous

@Preetha

- anonymous

@pooja195

- anonymous

@dan815

- anonymous

Not really sure what the first one is asking for, but I can help you with the second one.

- anonymous

That sound good?

- 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

Here's a nice little cheat-sheet for trig formulas: http://learnix.net/wordpress/wp-content/uploads/Trig-Cheat-Sheet-1.4.pdf

- anonymous

ok

- anonymous

i do not have arc lenght

- anonymous

Ugh, this is what I get for trying to do math while sleep deprived. Give me a minute, sorry.

- anonymous

lol its ok

- mathstudent55

Let's look at the first problem.
You are given an angular velocity, and you asked to find a linear velocity.

- anonymous

yes

- anonymous

are you still there

- 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

\(\ \large v = r\omega\)
In your case, you have the angular velocity given in revolutions per minute. You need to convert revolutions into radians.

- mathstudent55

Since the answer must be in miles per hour, you will also need to convert minutes into hours and inches into miles.

- anonymous

so angular velocity is 6500

- anonymous

i dont get it, what do i do next

- anonymous

@mathstudent55

- mathstudent55

We start with this:
\(\large v = r \omega = 7.25 ~in. \times ~3250 ~\dfrac{rev}{min}\)

- mathstudent55

Now we need to apply all the conversion factors.

- anonymous

but isnt angular velocity 6500

- 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

ohhhhh

- 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

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

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

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

What do you get?
\(v = \dfrac{7.25 \times 3250 \times 2 \times \pi \times 60}{12 \times 5280} \dfrac{miles}{hour} \)

- anonymous

one sec

- anonymous

140.2

- mathstudent55

That's what I got.

- anonymous

140.2 miles per hour

- anonymous

OK @LegendarySadist

- anonymous

so whats my first step

- 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

so arc length is 20/3

- anonymous

Correct. Now we can plug that into our other formula to find \(\large \sf \theta\) \[\large \sf \theta~=~\frac{Arc~length}{Radius}\]

- anonymous

bit i need theta in radians

- anonymous

but*

- anonymous

This is the solution in radians. That's why I chose this formula instead of the degrees formula.

- anonymous

i got 20/27

- anonymous

That would be it.

- 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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