anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@LexiLuvv2431 The true statement in that a tangent is positive in quadrant I.
anonymous
  • anonymous
tangent is positive in first quadrant
Nnesha
  • Nnesha
|dw:1439494300583:dw| remember the \(\huge\color{green}{\rm CAST }\) rule

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anonymous
  • anonymous
Thank you guys so much I have a few more can you guys help?
Nnesha
  • Nnesha
|dw:1439494347347:dw|
Nnesha
  • Nnesha
|dw:1439494382810:dw|
anonymous
  • anonymous
@LexiLuvv2431 u have ta give me da medal because I answerd it 1 lol
anonymous
  • anonymous
Find the sine, cosine, and tangent of 120 degrees. and I did lol
Nnesha
  • Nnesha
use unit circle (x,y) where x -coordinate = cos y-coordinate = sin \[\rm tan \theta = \frac{ \sin \theta }{ \cos \theta }\]
Nnesha
  • Nnesha
|dw:1439494649288:dw|
Nnesha
  • Nnesha
so cos= ? sin=? tan = ?
anonymous
  • anonymous
Sin is \[\frac{ \sqrt{3} }{ 2 }\]
anonymous
  • anonymous
Idk the others though
Nnesha
  • 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
  • anonymous
waittt okay so then would the cos equal 1/2 and the tan equal \[\sqrt{3}\] ?
Nnesha
  • Nnesha
cos is not just 1/2 look at first drawing i posted above cos is negative in which quadrant ?
anonymous
  • anonymous
Quadrant S
Nnesha
  • Nnesha
|dw:1439495263173:dw| quadrant numbers i guess you don't understand why i wrote CAST C= cos S=sin T=tan A= all functions
Nnesha
  • Nnesha
|dw:1439495345327:dw| cos is negative in 2nd and 3rd quadrant
Nnesha
  • Nnesha
now look at the unit circle 120 degree is in 2nd quadrant where cos value should be negative |dw:1439495460867:dw|
anonymous
  • anonymous
Noo sorry this is soo confusing for me.
Nnesha
  • Nnesha
just focus on it it's not hard you don't have to solve for anything just read the unit circle
anonymous
  • anonymous
Ohhhh I seee. I get it nowww
anonymous
  • anonymous
One more similar to this, The point (−3, 1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.
Nnesha
  • Nnesha
|dw:1439495526308:dw| it's like simple (x,y) graph
Nnesha
  • Nnesha
(x,y) x= cos y=sin so what is sin and cos in that equation ?
anonymous
  • anonymous
-3 and 1??
Nnesha
  • Nnesha
yew which one is cos
anonymous
  • anonymous
-3
Nnesha
  • 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
  • Nnesha
cos = negative (-3) and sin =positive (1) hint!
anonymous
  • anonymous
Quadrant 2
Nnesha
  • Nnesha
yep right !!
Nnesha
  • Nnesha
|dw:1439495963007:dw| apply Pythagorean theorem to find 3rd side of right triangle \[\huge\rm a^2+b^2=c^2\]
Nnesha
  • Nnesha
a=1 b=-3 solve for c
anonymous
  • anonymous
100?
Nnesha
  • Nnesha
hmm 100 no how did you get that ?
Nnesha
  • 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
  • anonymous
ohhh I forgot the 3 was negative. So -64
Nnesha
  • Nnesha
show your work how did you get that ?
anonymous
  • 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
  • Nnesha
okay \[(1)+(-3)^2=c^2\] when you take even power of negative you will get positive answer
Nnesha
  • Nnesha
and i just noticed you said 100 earlier
Nnesha
  • 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
  • Nnesha
wait no nvm
anonymous
  • anonymous
Okay now I'm confused
Nnesha
  • 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
  • anonymous
10 squared is 100
Nnesha
  • Nnesha
square root not square
Nnesha
  • 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
  • anonymous
Okay I have no idea how to do this part
Nnesha
  • Nnesha
\[\huge\rm \sqrt{c^2}= \sqrt{10}\]take square root both sides sqrt{c^2} = c so \[c =\sqrt{10}\]
Nnesha
  • Nnesha
|dw:1439497283978:dw|
anonymous
  • anonymous
Okayy I see
Nnesha
  • 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
  • Nnesha
|dw:1439497367283:dw| opp= opposite side of theta adj= adjacent of theta hyp=hypotenuse of theta
anonymous
  • anonymous
|dw:1439508247714:dw|
anonymous
  • anonymous
So is the answer finished?
Nnesha
  • Nnesha
yes right just write value of sin cosine and tan
anonymous
  • anonymous
How do I do that?
Nnesha
  • 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
  • Nnesha
|dw:1439498158642:dw|
anonymous
  • anonymous
Thank youu so much
Nnesha
  • 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
  • Nnesha
my pleasure:)

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