# Choose the point on the terminal side of 0. 0=2\pi/3 a) (1,-1) b)

Choose the point on the terminal side of 0.
$$\displaystyle{0}={2}\frac{\pi}{{3}}$$
a) $$\displaystyle{\left({1},-{1}\right)}$$
b) $$\displaystyle{\left({1},\sqrt{{{3}}}\right)}$$
c) $$\displaystyle{\left(-\sqrt{{{3}}},{1}\right)}$$

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Wasither1957
As we know from the Trigonpmetric functions
$$\displaystyle{\tan{\theta}}={\frac{{{y}}}{{{x}}}}$$
$$\displaystyle{\tan{{\frac{{{2}\pi}}{{{3}}}}}}={\frac{{{y}}}{{{x}}}}$$
$$\displaystyle-\sqrt{{{3}}}={\frac{{{y}}}{{{x}}}}$$
Now check the points whose $$\displaystyle{\frac{{{y}}}{{{x}}}}$$ value is equal to $$\displaystyle-\sqrt{{{3}}}$$. Then we can say that the point is on the terminal side of $$\displaystyle\theta$$
a) For the point (-1,1)
$$\displaystyle{\frac{{{y}}}{{{x}}}}={\frac{{{1}}}{{-{1}}}}=-{1}$$
b) For the point $$\displaystyle{\left({1},\sqrt{{{3}}}\right)}$$
$$\displaystyle{\frac{{{y}}}{{{x}}}}={\frac{{\sqrt{{{3}}}}}{{-{1}}}}=-\sqrt{{{3}}}$$
c) For the point $$\displaystyle{\left(-\sqrt{{{3}}},{1}\right)}$$
$$\displaystyle{\frac{{{y}}}{{{x}}}}={\frac{{{1}}}{{-\sqrt{{{3}}}}}}=-{\frac{{{1}}}{{\sqrt{{{3}}}}}}$$
So the point b is the terminal of $$\displaystyle\theta$$.