Find the Tangent Given the Point (1/5,-(2 square root of 6)/5)

$(\frac{1}{5}, - \frac{2 \sqrt{6}}{5})$

To find the $\tan (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(\frac{1}{5}, - \frac{2 \sqrt{6}}{5})$ , draw the triangle between the three points $(0, 0)$ , $(\frac{1}{5}, 0)$ , and $(\frac{1}{5}, - \frac{2 \sqrt{6}}{5})$ .

Opposite : $- \frac{2 \sqrt{6}}{5}$

Adjacent : $\frac{1}{5}$

$\tan (θ) = \frac{Opposite}{Adjacent}$ therefore $\tan (θ) = \frac{- \frac{2 \sqrt{6}}{5}}{\frac{1}{5}}$ .

$\frac{- \frac{2 \sqrt{6}}{5}}{\frac{1}{5}}$

Simplify $\tan (θ)$ .

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Multiply the numerator by the reciprocal of the denominator.

$\tan (θ) = - \frac{2 \sqrt{6}}{5} \cdot 5$

Cancel the common factor of $5$ .

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Move the leading negative in $- \frac{2 \sqrt{6}}{5}$ into the numerator.

$\tan (θ) = \frac{- 2 \sqrt{6}}{5} \cdot 5$

Cancel the common factor.

$\tan (θ) = \frac{- 2 \sqrt{6}}{5} \cdot 5$

Rewrite the expression.

$\tan (θ) = - 2 \sqrt{6}$

Approximate the result.

$\tan (θ) = - 2 \sqrt{6} \approx - 4.89897948$

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