Respuesta :

Solution

For this case we have the following angle:

[tex]\theta=-\frac{23\pi}{6}=-690º[/tex]

Part a:

Quadrant I

Part b:

We can do this 2*360 -690 = 30º

[tex]\frac{\pi}{6}=30º[/tex]

Part c

for this case we can do this:

[tex]x=1\cdot\cos 30=\frac{\sqrt[]{3}}{2},y=1\cdot\sin 30=\frac{1}{2}[/tex]

Then the point where theta intersect the unit circle is:

[tex](\frac{\sqrt[]{3}}{2},\frac{1}{2})[/tex]

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