Find the angle in radians and degrees for the point $left(-frac12,-fracsqrt32ight)$ on the unit circle

We need to find the angle corresponding to the point $(-\frac{1}{2},-\frac{\sqrt{3}}{2})$ on the unit circle. This point lies in the third quadrant where both sine and cosine are negative. The reference angle is given by:

$\text{Reference angle}=\arccos \left(\frac{1}{2}\right)=\frac{\pi }{3}$

Since the point is in the third quadrant, the angle in radians is:

$\theta =\pi +\frac{\pi }{3}=\frac{4\pi }{3}$

To convert this to degrees:

$\theta =\frac{4\pi }{3}\times \frac{180}{\pi }={240}^{\circ }$

Hence, the angle is $\frac{4\pi }{3}$ radians or ${240}^{\circ }$.

To find the angle for the point $left(-frac12,-fracsqrt32ight)$, observe that it lies in the third quadrant with a reference angle:

$extReferenceangle=arccosleft(frac12ight)=fracpi3$

The actual angle in radians is:

$heta=pi+fracpi3=frac4pi3$

In degrees:

$heta={240}^{circ}$

For the point $left(-frac12,-fracsqrt32ight)$, the angle is:

$heta=frac4pi3extradiansor{240}^{circ}$

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