# Choose the point on the terminal side of 0. 0=-60^{\circ} (a)(-1,-1)(b)(1

Choose the point on the terminal side of 0.
$$\displaystyle{0}=-{60}^{{\circ}}{\left({a}\right)}{\left(-{1},-{1}\right)}{\left({b}\right)}{\left({1},-\sqrt{{{3}}}\right)}{\left({c}\right)}{\left(-\sqrt{{{3}}},{1}\right)}$$

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Nathanael Webber
As we know from the Trigonometric functions
$$\displaystyle{\tan{\theta}}={\frac{{{y}}}{{{x}}}}$$
$$\displaystyle{\tan{-}}{60}^{{\circ}}={\frac{{{y}}}{{{x}}}}$$
$$\displaystyle-\sqrt{{{3}}}={\frac{{{y}}}{{{x}}}}$$
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 side of $$\displaystyle\theta$$.
Results:
B) $$\displaystyle{\left({1},-\sqrt{{{3}}}\right)}$$