Find the Cosine Given the Point (9/41,40/41)

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

Step 1

To find the $\cos (θ)$ 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 : $\frac{9}{41}$

Step 2

Find the hypotenuse using Pythagorean theorem $c = \sqrt{a^{2} + b^{2}}$ .

Apply the product rule to $\frac{9}{41}$ .

$\sqrt{\frac{9^{2}}{41^{2}} + {(\frac{40}{41})}^{2}}$

Raise $9$ to the power of $2$ .

$\sqrt{\frac{81}{41^{2}} + {(\frac{40}{41})}^{2}}$

Raise $41$ to the power of $2$ .

$\sqrt{\frac{81}{1681} + {(\frac{40}{41})}^{2}}$

Apply the product rule to $\frac{40}{41}$ .

$\sqrt{\frac{81}{1681} + \frac{40^{2}}{41^{2}}}$

Raise $40$ to the power of $2$ .

$\sqrt{\frac{81}{1681} + \frac{1600}{41^{2}}}$

Raise $41$ to the power of $2$ .

$\sqrt{\frac{81}{1681} + \frac{1600}{1681}}$

Combine the numerators over the common denominator.

$\sqrt{\frac{81 + 1600}{1681}}$

Add $81$ and $1600$ .

$\sqrt{\frac{1681}{1681}}$

Divide $1681$ by $1681$ .

$\sqrt{1}$

Any root of $1$ is $1$ .

$1$

Step 3

$\cos (θ) = \frac{Adjacent}{Hypotenuse}$ therefore $\cos (θ) = \frac{\frac{9}{41}}{1}$ .

$\frac{\frac{9}{41}}{1}$

Step 4

Divide $\frac{9}{41}$ by $1$ .

$\cos (θ) = \frac{9}{41}$

Step 5

Approximate the result.

$\cos (θ) = \frac{9}{41} \approx 0.\overline{21951}$

Find the Cosine Given the Point (9/41,40/41)

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