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Amenah8
 one year ago
FIND FAETA: cos faeta = (15/17), sin faeta > 0
Amenah8
 one year ago
FIND FAETA: cos faeta = (15/17), sin faeta > 0

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2do you mean cos(\theta) and sin(\theta), right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the requested angle \theta lies in the second quadrant, since we have this drawing: dw:1439224267236:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439224370587:dw

Amenah8
 one year ago
Best ResponseYou've already chosen the best response.0how did you know it was the 2nd quadrant?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2sionce if the point P, namely the second ends of your arc, lies in the second quadrant, then its xcoordinate, which is equal to cos(\theta) is negative, whereas its ycoordinate, which is sin(\theta) is positive, as requested from your problem

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so you have to find an angle whose measure is greater than 90 degrees and less than 180 degrees, whose cosine function is equal to 15/17

Amenah8
 one year ago
Best ResponseYou've already chosen the best response.0and i would use the graphing circle

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2for computation, you can use, for example, windows calculator

Amenah8
 one year ago
Best ResponseYou've already chosen the best response.0but if cos is just x, x=15/7. so then how do we know y (sin) is positive? couldn't it have been in the 3rd quadrant?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2sin(\theta) has to be positive, since it is a data of your problem

Amenah8
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean by a data of my problem?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439224865101:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2as we can see from my last drawing, only points which belongs to the first quadrant and second quadrant, have ycoordinate >0

Amenah8
 one year ago
Best ResponseYou've already chosen the best response.0oh, because in third and 4th quadrants, y is negative!

Amenah8
 one year ago
Best ResponseYou've already chosen the best response.0so how would i solve for theta? on the unit circle, the only x "coordinates" are the negative square root of 3 divided by 2, 1/2, and the negative square root of 2 divided by 2. None of those exactly equal 15/17.

Amenah8
 one year ago
Best ResponseYou've already chosen the best response.0but \[\sqrt{3}/2\] is close

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2you have to find \theta, right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2if cos(\theta)= 15/17, then using windows calculator I got: \theta= 151.93 degrees, namely \theta= 152 degrees

Amenah8
 one year ago
Best ResponseYou've already chosen the best response.0how did you get 151.93?
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