@mathstudent55 could you help me with this one as well? Thanks in advance- cos θ = 1/2 and θ terminates in QIV. So far I have this: http://prntscr.com/8dhn18 \[\text{not sure how the answer is not }\sqrt{3}\text{ as I ran through the calculations several times}\]\[\text{with the same result}\]

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@mathstudent55 could you help me with this one as well? Thanks in advance- cos θ = 1/2 and θ terminates in QIV. So far I have this: http://prntscr.com/8dhn18 \[\text{not sure how the answer is not }\sqrt{3}\text{ as I ran through the calculations several times}\]\[\text{with the same result}\]

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\[y=\sqrt{2^{2}-1^{2}}=\sqrt{4-1}=\sqrt{3}???\]
What is y?
0.5 ?! Wait what?! o_o
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I thought this was a 30 degree angle so I went off that, and I double checked in case.... but apparently I had the wrong idea ...Isn't cos x/r?
Your problem does state that cos(theta) = 1/2, right?
You have the right idea. It's a 30-60-90 triangle. There is a 30 deg angle and a 60 deg angle.
Yes... and I got lost afterwards since the answer is not the 'obvious' root 3. e_e
In this case, we are dealing witht he 60-deg angle.
I thought it was the 30 degree... (feels stupid) x_x
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Oh. Yes.
I neglected the quadrant. Sorry e_e
Now notice that the y-coordinate is negative, so sine and cosecant will be negative. Since the x-coordinate is positive, the cosine and secant are positive. The tangent and cotangent are negative.
I think my issues were due to neglecting to check which quadrant I was in... I will make sure, brb
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Yup, got them right this time. Thanks for correcting my stupid mistake ^^;
You're welcome.

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