Find the Tangent (9/41,40/41)

$(\frac{9}{41}, \frac{40}{41})$

To find the $\tan (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(\frac{9}{41}, \frac{40}{41})$ , draw the triangle between the three points $(0, 0)$ , $(\frac{9}{41}, 0)$ , and $(\frac{9}{41}, \frac{40}{41})$ .

Opposite : $\frac{40}{41}$

Adjacent : $\frac{9}{41}$

$\tan (θ) = \frac{Opposite}{Adjacent}$ therefore $\tan (θ) = \frac{\frac{40}{41}}{\frac{9}{41}}$ .

$\frac{\frac{40}{41}}{\frac{9}{41}}$

Simplify $\tan (θ)$ .

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Multiply the numerator by the reciprocal of the denominator.

$\tan (θ) = \frac{40}{41} \cdot \frac{41}{9}$

Cancel the common factor of $41$ .

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Cancel the common factor.

$\tan (θ) = \frac{40}{41} \cdot \frac{41}{9}$

Rewrite the expression.

$\tan (θ) = 40 (\frac{1}{9})$

Combine $40$ and $\frac{1}{9}$ .

$\tan (θ) = \frac{40}{9}$

Approximate the result.

$\tan (θ) = \frac{40}{9} \approx 4.\overline{4}$

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