# How do you find the exact value for tan 120?

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30
Jan 21, 2018

$- \sqrt{3}$

#### Explanation:

First of all $\tan \left(\angle\right) = \left(\sin \frac{\angle}{\cos} \left(\angle\right)\right)$
So $\tan \left(120\right) = \sin \frac{120}{\cos} \left(120\right)$
$\sin \left(120\right) = \frac{\sqrt{3}}{2} \mathmr{and} \cos \left(120\right) = - \frac{1}{2}$
therefore $\tan \left(120\right) = \frac{\frac{\sqrt{3}}{2}}{- \frac{1}{2}}$
canceling out the $\frac{1}{2}$ from the numerator and the denominator we get
$- \sqrt{3}$
but remember $\sqrt{3}$ doesn't has any rational value .

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21
Jan 1, 2016

$\tan {120}^{\circ} = - \sqrt{3}$

#### Explanation:

$\tan {120}^{\circ} = \sin {120}^{\circ} / \cos {120}^{\circ}$

Let's evaluate $\sin {120}^{\circ}$ first

$\sin {120}^{\circ} = \sin 2 \left({60}^{\circ}\right)$
$\sin {120}^{\circ} = 2 \sin {60}^{\circ} \cos {60}^{\circ}$
$\sin {120}^{\circ} = 2 \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right)$
$\sin {120}^{\circ} = \frac{\sqrt{3}}{2}$

Since we found the value of $\sin {120}^{\circ}$, we can say that

$\cos {120}^{\circ} = - \frac{1}{2}$

Because ${120}^{\circ}$ is in Quadrant 2.

Since $\tan \theta = \sin \frac{\theta}{\cos} \theta , \frac{\sqrt{3}}{2} \cdot 2$

(Dividing by 1/2 can be evaluated by doubling (multiplying by 2))

$\therefore \tan {120}^{\circ} = - \sqrt{3}$

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11
Sep 18, 2017

$\tan \left(120\right) = - \sqrt{3}$

#### Explanation:

$\tan \left(120\right) = \tan \left(2 \cdot 60\right)$

We know that

$\tan \left(2 A\right) = \frac{2 \tan \left(A\right)}{1 - {\tan}^{2} \left(A\right)}$

hence

$\tan \left(2 \cdot 60\right) = \frac{2 \tan \left(60\right)}{1 - {\tan}^{2} \left(60\right)}$

Therefore,

$\tan \left(120\right) = \frac{2 \sqrt{3}}{1 - {\left(\sqrt{3}\right)}^{2}}$

$= \frac{2 \sqrt{3}}{1 - 3}$

$= \frac{2 \sqrt{3}}{- 2}$

$= - \sqrt{3}$

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