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anzhelika [568]
1 year ago
14

LOTS OF POUNT PLS HELP

Mathematics
1 answer:
weqwewe [10]1 year ago
4 0

Answer:

Option 2

Step-by-step explanation:

Using the quadrstic formula,

\tan \theta=\frac{-2 \pm \sqrt{2^2 - 4(\sqrt{3})(-\sqrt{3})}}{2\sqrt{3}} \\ \\ =\frac{-2 \pm 4}{2\sqrt{3}} \\ \\ =\frac{-1 \pm 2}{\sqrt{3}} \\ \\ = -\sqrt{3}, \frac{1}{\sqrt{3}}

<u>Case</u><u> </u><u>1</u>

\tan \theta=-\sqrt{3} \implies \theta=\frac{2\pi}{3}, \frac{5\pi}{3}

<u>Case</u><u> </u><u>2</u>

\tan \theta=\frac{1}{\sqrt{3}} \implies \theta=\frac{\pi}{6}, \frac{7\pi}{6}

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