If tan theta = - frac{1}{sqrt{3}},sin theta = | vedclass

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(B) Given that $	an 	heta = -\frac{1}{\sqrt{3}}$,$\sin 	heta = \frac{1}{2}$,and $\cos 	heta = -\frac{\sqrt{3}}{2}$.Since $\sin 	heta > 0$ and $\c

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$\frac{5\pi}{6}$

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(B) Given that $	an 	heta = -\frac{1}{\sqrt{3}}$,$\sin 	heta = \frac{1}{2}$,and $\cos 	heta = -\frac{\sqrt{3}}{2}$.Since $\sin 	heta > 0$ and $\cos 	heta The reference angle $\alpha$ is given by $	an \alpha = \frac{1}{\sqrt{3}}$,which implies $\alpha = \frac{\pi}{6}$.In the second quadrant,$	heta = \pi - \alpha = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.Thus,the principal value of $	heta$ is $\frac{5\pi}{6}$.

If $\tan \theta = - \frac{1}{\sqrt{3}}$,$\sin \theta = \frac{1}{2}$,and $\cos \theta = - \frac{\sqrt{3}}{2}$,then the principal value of $\theta$ is:

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