Suppose that angle $ heta $ is positioned on the unit circle such that $ heta = frac{5pi}{6} $. Determine the coordinates of the point where the terminal side of $ heta $ intersects the unit circle.

First, we identify that $ \theta = \frac{5\pi}{6} $ is in the second quadrant. The reference angle is $ \pi – \frac{5\pi}{6} = \frac{\pi}{6} $.

In the unit circle, the cosine and sine of $ \frac{\pi}{6} $ are $ \frac{\sqrt{3}}{2} $ and $ \frac{1}{2} $, respectively.

Therefore, in the second quadrant, the coordinates are $ (-\frac{\sqrt{3}}{2}, \frac{1}{2}) $.

$ \text{Coordinates: } \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) $

To solve for the coordinates, we use the unit circle properties. Given $ heta = frac{5pi}{6} $, we find the reference angle $ frac{pi}{6} $.

Using the unit circle, $ cos(frac{5pi}{6}) = -cos(frac{pi}{6}) = -frac{sqrt{3}}{2} $ and $ sin(frac{5pi}{6}) = sin(frac{pi}{6}) = frac{1}{2} $.

Hence, the coordinates are $ left(-frac{sqrt{3}}{2}, frac{1}{2}
ight) $.

$ ext{Coordinates: } left(-frac{sqrt{3}}{2}, frac{1}{2}
ight) $

For $ heta = frac{5pi}{6} $, the reference angle is $ frac{pi}{6} $.

In the second quadrant, $ cos(frac{pi}{6}) = frac{sqrt{3}}{2} $, so $ cos(frac{5pi}{6}) = -frac{sqrt{3}}{2} $.

Similarly, $ sin(frac{5pi}{6}) = frac{1}{2} $.

$ ext{Coordinates: } left(-frac{sqrt{3}}{2}, frac{1}{2}
ight) $

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