Need help with homework

Question

asked Jan 4, 2023 44.2k views

1 Answer

Dovy · Answer 1 · 2023-01-11T04:04:27+0000

Solution

For this case we have the following angle:

$\theta=-(23\pi)/(6)=-690º$

Part a:

Quadrant I

Part b:

We can do this 2*360 -690 = 30º

$(\pi)/(6)=30º$

Part c

for this case we can do this:

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

Then the point where theta intersect the unit circle is:

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