Which of the following is true?
Tangent is positive in Quadrant I.
Sine is negative in Quadrant II.
Cosine is positive in Quadrant III.
Sine is positive in Quadrant IV.

- anonymous

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- anonymous

@LexiLuvv2431 The true statement in that a tangent is positive in quadrant I.

- anonymous

tangent is positive in first quadrant

- Nnesha

|dw:1439494300583:dw|
remember the \(\huge\color{green}{\rm CAST }\) rule

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## More answers

- anonymous

Thank you guys so much I have a few more can you guys help?

- Nnesha

|dw:1439494347347:dw|

- Nnesha

|dw:1439494382810:dw|

- anonymous

@LexiLuvv2431 u have ta give me da medal because I answerd it 1 lol

- anonymous

Find the sine, cosine, and tangent of 120 degrees. and I did lol

- Nnesha

use unit circle
(x,y) where x -coordinate = cos
y-coordinate = sin
\[\rm tan \theta = \frac{ \sin \theta }{ \cos \theta }\]

- Nnesha

|dw:1439494649288:dw|

- Nnesha

so cos= ?
sin=?
tan = ?

- anonymous

Sin is \[\frac{ \sqrt{3} }{ 2 }\]

- anonymous

Idk the others though

- Nnesha

hm
yes that's right
so if sin =sqrt{3}/2
cos = what look at the same parentheses where x-coordinate represents cos

- anonymous

waittt okay so then would the cos equal 1/2 and the tan equal \[\sqrt{3}\] ?

- Nnesha

cos is not just 1/2 look at first drawing i posted above
cos is negative in which quadrant ?

- anonymous

Quadrant S

- Nnesha

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quadrant numbers
i guess you don't understand why i wrote CAST
C= cos
S=sin
T=tan
A= all functions

- Nnesha

|dw:1439495345327:dw|
cos is negative in 2nd and 3rd quadrant

- Nnesha

now look at the unit circle
120 degree is in 2nd quadrant where cos value should be negative |dw:1439495460867:dw|

- anonymous

Noo sorry this is soo confusing for me.

- Nnesha

just focus on it
it's not hard you don't have to solve for anything
just read the unit circle

- anonymous

Ohhhh I seee. I get it nowww

- anonymous

One more similar to this,
The point (âˆ’3, 1) is on the terminal side of angle Î¸, in standard position. What are the values of sine, cosine, and tangent of Î¸? Make sure to show all work.

- Nnesha

|dw:1439495526308:dw|
it's like simple (x,y) graph

- Nnesha

(x,y)
x= cos
y=sin
so what is sin and cos in that equation ?

- anonymous

-3 and 1??

- Nnesha

yew which one is cos

- anonymous

-3

- Nnesha

yes right so you have draw a right triangle
cos =-3 sin =1 sow at which quadrant you should draw a right triangle ?

- Nnesha

cos = negative (-3) and sin =positive (1) hint!

- anonymous

Quadrant 2

- Nnesha

yep right !!

- Nnesha

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apply Pythagorean theorem to find 3rd side of right triangle \[\huge\rm a^2+b^2=c^2\]

- Nnesha

a=1
b=-3
solve for c

- anonymous

100?

- Nnesha

hmm 100 no how did you get that ?

- Nnesha

replace a with 1 and b with -3 then take square of both numbers
and then add them
after that take square root

- anonymous

ohhh I forgot the 3 was negative.
So -64

- Nnesha

show your work how did you get that ?

- anonymous

I did 1*1 and got one and the -3 * -3 and got -9 and I added that and got -8 and then multiplied -8 * -8 and got -64

- Nnesha

okay \[(1)+(-3)^2=c^2\] when you take even power of negative you will get positive answer

- Nnesha

and i just noticed you said 100 earlier

- Nnesha

which is correct but i thought you r saying c =100
c^2=100
take square root both sides you need c not c^2

- Nnesha

wait no nvm

- anonymous

Okay now I'm confused

- Nnesha

\[\huge\rm (1)^2 + (-3)^2=c^2\]\[\rm 1+9=c^2\]
(-3 times -3= 9)
so now add 1+9\[\huge\rm 10=c^2\] now take square root both sides

- anonymous

10 squared is 100

- Nnesha

square root
not square

- Nnesha

to cancel out square you should take square root both sides
to cancel out square root you should take square both sides
so there is c^2 to cancel out square take square root

- anonymous

Okay I have no idea how to do this part

- Nnesha

\[\huge\rm \sqrt{c^2}= \sqrt{10}\]take square root both sides
sqrt{c^2} = c so \[c =\sqrt{10}\]

- Nnesha

|dw:1439497283978:dw|

- anonymous

Okayy I see

- Nnesha

now remember definition of sin,cos and tan
you need this \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

- Nnesha

|dw:1439497367283:dw|
opp= opposite side of theta
adj= adjacent of theta
hyp=hypotenuse of theta

- anonymous

|dw:1439508247714:dw|

- anonymous

So is the answer finished?

- Nnesha

yes right just write value of sin cosine and tan

- anonymous

How do I do that?

- Nnesha

\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]
just replace opposite , adjacent hyp words with their values

- Nnesha

|dw:1439498158642:dw|

- anonymous

Thank youu so much

- Nnesha

like sin = opposite over hyp
so opposite side of theta is 1
and hyp = sqrt{10}
write it as \[\sin \theta =\frac{ 1 }{ \sqrt{10} }\] you can't leave the radical sign at the denominator so multiply top and bottom by square root 10

- Nnesha

my pleasure:)

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