Find the Cosine Given the Point (48/73,55/73)

$(\frac{48}{73}, \frac{55}{73})$

Step 1

To find the $\cos (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(\frac{48}{73}, \frac{55}{73})$ , draw the triangle between the three points $(0, 0)$ , $(\frac{48}{73}, 0)$ , and $(\frac{48}{73}, \frac{55}{73})$ .

Opposite : $\frac{55}{73}$

Adjacent : $\frac{48}{73}$

Step 2

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

Apply the product rule to $\frac{48}{73}$ .

$\sqrt{\frac{48^{2}}{73^{2}} + {(\frac{55}{73})}^{2}}$

Raise $48$ to the power of $2$ .

$\sqrt{\frac{2304}{73^{2}} + {(\frac{55}{73})}^{2}}$

Raise $73$ to the power of $2$ .

$\sqrt{\frac{2304}{5329} + {(\frac{55}{73})}^{2}}$

Apply the product rule to $\frac{55}{73}$ .

$\sqrt{\frac{2304}{5329} + \frac{55^{2}}{73^{2}}}$

Raise $55$ to the power of $2$ .

$\sqrt{\frac{2304}{5329} + \frac{3025}{73^{2}}}$

Raise $73$ to the power of $2$ .

$\sqrt{\frac{2304}{5329} + \frac{3025}{5329}}$

Combine the numerators over the common denominator.

$\sqrt{\frac{2304 + 3025}{5329}}$

Add $2304$ and $3025$ .

$\sqrt{\frac{5329}{5329}}$

Divide $5329$ by $5329$ .

$\sqrt{1}$

Any root of $1$ is $1$ .

$1$

Step 3

$\cos (θ) = \frac{Adjacent}{Hypotenuse}$ therefore $\cos (θ) = \frac{\frac{48}{73}}{1}$ .

$\frac{\frac{48}{73}}{1}$

Step 4

Divide $\frac{48}{73}$ by $1$ .

$\cos (θ) = \frac{48}{73}$

Step 5

Approximate the result.

$\cos (θ) = \frac{48}{73} \approx 0.65753424$

Find the Cosine Given the Point (48/73,55/73)

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