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anonymous
 one year ago
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
 one year ago
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@LexiLuvv2431 The true statement in that a tangent is positive in quadrant I.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0tangent is positive in first quadrant

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439494300583:dw remember the \(\huge\color{green}{\rm CAST }\) rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you guys so much I have a few more can you guys help?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@LexiLuvv2431 u have ta give me da medal because I answerd it 1 lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find the sine, cosine, and tangent of 120 degrees. and I did lol

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0use unit circle (x,y) where x coordinate = cos ycoordinate = sin \[\rm tan \theta = \frac{ \sin \theta }{ \cos \theta }\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0so cos= ? sin=? tan = ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sin is \[\frac{ \sqrt{3} }{ 2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Idk the others though

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0hm yes that's right so if sin =sqrt{3}/2 cos = what look at the same parentheses where xcoordinate represents cos

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0waittt okay so then would the cos equal 1/2 and the tan equal \[\sqrt{3}\] ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0cos is not just 1/2 look at first drawing i posted above cos is negative in which quadrant ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439495263173:dw quadrant numbers i guess you don't understand why i wrote CAST C= cos S=sin T=tan A= all functions

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439495345327:dw cos is negative in 2nd and 3rd quadrant

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0now look at the unit circle 120 degree is in 2nd quadrant where cos value should be negative dw:1439495460867:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Noo sorry this is soo confusing for me.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0just focus on it it's not hard you don't have to solve for anything just read the unit circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhhh I seee. I get it nowww

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439495526308:dw it's like simple (x,y) graph

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0(x,y) x= cos y=sin so what is sin and cos in that equation ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0yes right so you have draw a right triangle cos =3 sin =1 sow at which quadrant you should draw a right triangle ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0cos = negative (3) and sin =positive (1) hint!

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439495963007:dw apply Pythagorean theorem to find 3rd side of right triangle \[\huge\rm a^2+b^2=c^2\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0hmm 100 no how did you get that ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0replace a with 1 and b with 3 then take square of both numbers and then add them after that take square root

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhh I forgot the 3 was negative. So 64

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0show your work how did you get that ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0okay \[(1)+(3)^2=c^2\] when you take even power of negative you will get positive answer

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0and i just noticed you said 100 earlier

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0which is correct but i thought you r saying c =100 c^2=100 take square root both sides you need c not c^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay now I'm confused

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge\rm (1)^2 + (3)^2=c^2\]\[\rm 1+9=c^2\] (3 times 3= 9) so now add 1+9\[\huge\rm 10=c^2\] now take square root both sides

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0to 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
 one year ago
Best ResponseYou've already chosen the best response.0Okay I have no idea how to do this part

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge\rm \sqrt{c^2}= \sqrt{10}\]take square root both sides sqrt{c^2} = c so \[c =\sqrt{10}\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0now 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439497367283:dw opp= opposite side of theta adj= adjacent of theta hyp=hypotenuse of theta

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439508247714:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So is the answer finished?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0yes right just write value of sin cosine and tan

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\] just replace opposite , adjacent hyp words with their values

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0like 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
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