Find the Cosine Given the Point (16/65,63/65)

$(\frac{16}{65}, \frac{63}{65})$

To find the $\cos (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(\frac{16}{65}, \frac{63}{65})$ , draw the triangle between the three points $(0, 0)$ , $(\frac{16}{65}, 0)$ , and $(\frac{16}{65}, \frac{63}{65})$ .

Opposite : $\frac{63}{65}$

Adjacent : $\frac{16}{65}$

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

Apply the product rule to $\frac{16}{65}$ .

$\sqrt{\frac{16^{2}}{65^{2}} + {(\frac{63}{65})}^{2}}$

Raise $16$ to the power of $2$ .

$\sqrt{\frac{256}{65^{2}} + {(\frac{63}{65})}^{2}}$

Raise $65$ to the power of $2$ .

$\sqrt{\frac{256}{4225} + {(\frac{63}{65})}^{2}}$

Apply the product rule to $\frac{63}{65}$ .

$\sqrt{\frac{256}{4225} + \frac{63^{2}}{65^{2}}}$

Raise $63$ to the power of $2$ .

$\sqrt{\frac{256}{4225} + \frac{3969}{65^{2}}}$

Raise $65$ to the power of $2$ .

$\sqrt{\frac{256}{4225} + \frac{3969}{4225}}$

Combine the numerators over the common denominator.

$\sqrt{\frac{256 + 3969}{4225}}$

Add $256$ and $3969$ .

$\sqrt{\frac{4225}{4225}}$

Divide $4225$ by $4225$ .

$\sqrt{1}$

Any root of $1$ is $1$ .

$1$

$\cos (θ) = \frac{Adjacent}{Hypotenuse}$ therefore $\cos (θ) = \frac{\frac{16}{65}}{1}$ .

$\frac{\frac{16}{65}}{1}$

Divide $\frac{16}{65}$ by $1$ .

$\cos (θ) = \frac{16}{65}$

Approximate the result.

$\cos (θ) = \frac{16}{65} \approx 0.2\overline{461538}$

Find the Cosine Given the Point (16/65,63/65)

Related Questions

Support LCMGCF.com by disabling your adblocker.