Find the Tangent Given the Point (60/61,-11/61)

$(\frac{60}{61}, - \frac{11}{61})$

To find the $\tan (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(\frac{60}{61}, - \frac{11}{61})$ , draw the triangle between the three points $(0, 0)$ , $(\frac{60}{61}, 0)$ , and $(\frac{60}{61}, - \frac{11}{61})$ .

Opposite : $- \frac{11}{61}$

Adjacent : $\frac{60}{61}$

$\tan (θ) = \frac{Opposite}{Adjacent}$ therefore $\tan (θ) = \frac{- \frac{11}{61}}{\frac{60}{61}}$ .

$\frac{- \frac{11}{61}}{\frac{60}{61}}$

Simplify $\tan (θ)$ .

Multiply the numerator by the reciprocal of the denominator.

$\tan (θ) = - \frac{11}{61} \cdot \frac{61}{60}$

Cancel the common factor of $61$ .

Move the leading negative in $- \frac{11}{61}$ into the numerator.

$\tan (θ) = \frac{- 11}{61} \cdot \frac{61}{60}$

Cancel the common factor.

$\tan (θ) = \frac{- 11}{61} \cdot \frac{61}{60}$

Rewrite the expression.

$\tan (θ) = - 11 (\frac{1}{60})$

Combine $- 11$ and $\frac{1}{60}$ .

$\tan (θ) = \frac{- 11}{60}$

Move the negative in front of the fraction.

$\tan (θ) = - \frac{11}{60}$

Approximate the result.

$\tan (θ) = - \frac{11}{60} \approx - 0.18\overline{3}$

Find the Tangent Given the Point (60/61,-11/61)

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