How do you evaluate "cot240#?

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How do you evaluate "cot240#?

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Answer 1 · 2015-11-29T22:09:06

$\frac{\sqrt{3}}{3}$ $sqrt(3)/3$

Explanation:

$cot (240^{\circ})$ $cot(240^@)$

$= \frac{1}{tan (240^{\circ})}$ $=1/tan(240^@)$

$= \frac{1}{\sqrt{3}}$ $=1/sqrt(3)$

$= \frac{1}{1} \div \frac{\sqrt{3}}{1}$ $=1/1-:sqrt(3)/1$

$= \frac{1}{1} \cdot \frac{1}{\sqrt{3}}$ $=1/1*1/sqrt(3)$

$= \frac{1}{\sqrt{3}}$ $=1/sqrt(3)$

$= \frac{1}{\sqrt{3}} \cdot (\frac{\sqrt{3}}{\sqrt{3}}) \Leftarrow$ $=1/sqrt(3) * (sqrt(3)/sqrt(3)) lArr$ rationalize

$= \frac{\sqrt{3}}{3}$ $=sqrt(3)/3$

Note:
Whenever you have a fraction where the denominator is a square root, you must rationalize the fraction. This is done by multiplying the fraction with another fraction with the numerator and denominator being the same as the square root denominator.

For example:

$\frac{5}{\sqrt{6}}$ $5/sqrt(6)$

$= \frac{5}{\sqrt{6}} \cdot (\frac{\sqrt{6}}{\sqrt{6}})$ $=5/sqrt(6) * (sqrt(6)/sqrt(6))$

$= \frac{5 \sqrt{6}}{6}$ $=(5sqrt(6))/6$

Answer 2 · 2015-11-29T23:07:40

Find cot 240

Ans: $- \frac{\sqrt{3}}{3}$ $-sqrt3/3$

Explanation:

Trig table of special arcs and unit circle -->
$cot (240) = cot (60 + 180) = cot (60) = \frac{\sqrt{3}}{3}$ $cot (240) = cot (60 + 180) = cot (60) = sqrt3/3$

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How do you evaluate "cot240#?

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