# How do you calculate \tan(\arccos(\frac{5}{13})) ?

How do you calculate $\mathrm{tan}\left(\mathrm{arccos}\left(\frac{5}{13}\right)\right)$ ?
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Amina Hall
If you set $\alpha =\mathrm{arccos}\left(\frac{5}{13}\right)⇒\mathrm{cos}\alpha =\frac{5}{13}$ than we have to calculate:
$\mathrm{tan}\alpha =\frac{\mathrm{sin}\alpha }{\mathrm{cos}\alpha }$
$\mathrm{cos}\alpha =\frac{5}{13}⇒\mathrm{sin}\alpha =\sqrt{1-{\mathrm{cos}}^{2}\alpha }=\sqrt{1-\frac{25}{169}}$
$=\sqrt{\frac{169-25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$
So:
$\mathrm{tan}\alpha =\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{13}\cdot \frac{13}{5}=\frac{12}{5}$
###### Not exactly what you’re looking for?
Aaron Hughes
$\mathrm{tan}\left(\mathrm{arccos}\left(\frac{5}{13}\right)\right)$
Draw a triangle in the plane with vertices $\left(\frac{5}{13},\sqrt{{1}^{2}-{\left(\frac{5}{13}\right)}^{2}}\right),\left(\frac{5}{13},0\right)$, and the origin. Then $\left(\frac{5}{13}\right)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $\left(\frac{5}{13},\sqrt{{1}^{2}-{\left(\frac{5}{13}\right)}^{2}}\right)$. Therefore, $\mathrm{tan}\left(\mathrm{arccos}\left(\frac{5}{13}\right)\right)$ is $\frac{\sqrt{{1}^{2}-{\left(\frac{5}{13}\right)}^{2}}}{\frac{5}{13}}$.
Multiply the numerator by the reciprocal of the denominator.
$\sqrt{{1}^{2}-{\left(\frac{5}{13}\right)}^{2}}\frac{13}{5}$
One to any power is one.
$\sqrt{1-{\left(\frac{5}{13}\right)}^{2}}\frac{13}{5}$
Apply the product rule to $\frac{5}{13}$
Raise 5 to the power of 2.
$\sqrt{1-\frac{25}{{13}^{2}}}\frac{13}{5}$
Raise 13 to the power of 2.
$\sqrt{1-\frac{25}{169}}\frac{13}{5}$
Write 1 as a fraction with a common denomiantor.
$\sqrt{\frac{169}{169}-\frac{25}{169}}\frac{13}{5}$
Combine the numerators over the common denominator.
$\sqrt{\frac{169-25}{169}}\frac{13}{5}$
Subtract 25 from 169.
$\sqrt{\frac{144}{169}}\frac{13}{5}$
Rewrite $\sqrt{\frac{144}{169}}$ as $\frac{\sqrt{144}}{\sqrt{169}}$.
$\frac{\sqrt{144}}{\sqrt{169}}\cdot \frac{13}{5}$
Simplify the numerator.
$\frac{12}{\sqrt{169}}\cdot \frac{13}{5}$
Simplify the denomiantor.
$\frac{12}{13}\cdot \frac{13}{5}$
Simplify terms.
$\frac{12}{5}$
The result can be shown in multiple forms.
Exact Form:
$\frac{12}{5}$
Decimal Form:
$2.4$
Mixed Number Form: $2\frac{2}{5}$