Question

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

Answer 1

@Nnesha

Answer 2

Pls help

Answer 3

@Preetha

Answer 4

@pooja195

Answer 5

@dan815

Answer 6

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

Answer 7

That sound good?

Answer 8

Well I mean, it's really simple. The angle \(\theta\) can be found using the formula \[\large \sf \theta~=~\frac{arc~length}{radius}\]

Answer 9

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

Answer 10

ok

Answer 11

i do not have arc lenght

Answer 12

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

Answer 13

lol its ok

Answer 14

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

Answer 15

yes

Answer 16

are you still there

Answer 17

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.

Answer 18

\(\ \large v = r\omega\)

In your case, you have the angular velocity given in revolutions per minute. You need to convert revolutions into radians.

Answer 19

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

Answer 20

so angular velocity is 6500

Answer 21

i dont get it, what do i do next

Answer 22

@mathstudent55

Answer 23

We start with this:

\(\large v = r \omega = 7.25 ~in. \times ~3250 ~\dfrac{rev}{min}\)

Answer 24

Now we need to apply all the conversion factors.

Answer 25

but isnt angular velocity 6500

Answer 26

Angular velocity needs to be in radians per minute.
You have revolutions per minute.
1 revolution is \(2 \pi\) radians, not 2 radians.

Answer 27

ohhhhh

Answer 28

Let's deal with the conversion of revolutions to radians. It's in red below.

\(\large v = r \omega = 7.25 ~in. \times ~3250 ~\dfrac{rev}{min} \color{red}{\times \dfrac{2 \pi ~rad }{1~rev}}\)

Answer 29

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. \[\large \sf \theta~=~\frac{Arc~length}{Radius}\]

Answer 30

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.

Answer 31

The conversion from minutes to hours is in blue below.
60 minutes = 1 hour

\(v = r \omega = 7.25 ~in. \color{green}{\times \dfrac{1~ft}{12~in.} \times \dfrac{1~mile}{5280~ft} } \times ~3250 ~\dfrac{rev}{min} \color{red}{\times \dfrac{2 \pi ~rad }{1~rev}}
\color{blue} {\times \dfrac{60~min}{1~hour} }\)

Answer 32

What do you get?

\(v = \dfrac{7.25 \times 3250 \times 2 \times \pi \times 60}{12 \times 5280} \dfrac{miles}{hour} \)

Answer 33

one sec

Answer 34

140.2

Answer 35

That's what I got.

Answer 36

140.2 miles per hour

Answer 37

OK @LegendarySadist

Answer 38

so whats my first step

Answer 39

That would be to find the arc length \[\large \sf Area~=~\frac{1}{2} \times r \times Arc \huge \rightarrow \large \frac{2 \times Area}{r}~=~Arc\]

Answer 40

so arc length is 20/3

Answer 41

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

Answer 42

bit i need theta in radians

Answer 43

but*

Answer 44

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

Answer 45

i got 20/27

Answer 46

That would be it.

Answer 47

Second problem:

\(\Large A_{sector} = \dfrac{\theta}{2 \pi ~rad}\pi r^2\)


\(\Large \theta = \dfrac{2 \pi A}{\pi r^2} = \dfrac{2A}{r^2} = \dfrac{2 \times 30 ft^2}{(9~ft)^2} = \dfrac{60}{81} = \dfrac{20}{27}\)