Trig question

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BaseballB6236

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Let pheta be an angle in quadrant IV, such that cot pheta = ( - sqrt 5) /2 .

Find sin pheta and sec pheta by using fundamental trigonometric identities.

I'm not sure what I'm doing wrong but, don't you assume because cot pheta equals that adjacent = -sqrt 5 and opposite = 2

now you do pythagorean to find hypo... and that's sqrt 9


so sin pheta = 2/ sqrt 9 and sec pheta = ??
 
BaseballB6236 said:
don't you assume because cot pheta equals that adjacent = -sqrt 5 and opposite = 2

No, you do not assume. Draw a picture, if you need to.

In Quadrant IV, adjacent = sqrt(5) and opposite = -2

find hypo... and that's sqrt 9

Why do you not simplify sqrt(9) ?

Then, try again with the correct signs.
Click to expand...
 
No, you do not assume. Draw a picture, if you need to.

In Quadrant IV, adjacent = sqrt(5) and opposite = -2

hypo = 3

sin pheta = -2/3

sec pheta = 1/ (sqrt 5/ 3) = 3/ sqrt 5
 
The thing I don't understand though is if it's in quadrant IV, y coords are negative, x coords are positive..how does that effect the hypo, adjacent, opposite? How do you know when the hypo, adjacent, opposite are negative are positive?
 


The hypotenuse is never negative.

Why do you believe that x-coordinates are negative in Quadrant IV ?

 
Re:

mmm4444bot said:


The hypotenuse is never negative.

Why do you believe that x-coordinates are negative in Quadrant IV ?
Click to expand...

It was a typo. Ok hypotenuse never being negative makes things clearer now. In this case the opposite is the y and adjacent is the x


Thanks for your help
 
BaseballB6236 said:
sin pheta = -2/3 Correct

sec pheta = 3/sqrt 5 Correct

pheta = 0.7297276563 radians or 318.1896851 degrees, approximated
Click to expand...

As for determining the signs, we go by the coordinates of the point where the terminal ray of the angle intersects the circle with radius = hypotenuse.

 
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