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

vestirme4 2021-10-30 Answered
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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Expert Answer

Nathanael Webber
Answered 2021-10-31 Author has 12156 answers
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) \(\displaystyle{\left({1},-\sqrt{{{3}}}\right)}\)
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