blitzbabeiy

## Answered question

2022-01-21

How do you calculate $\mathrm{tan}\left(\mathrm{arccos}\left(\frac{5}{13}\right)\right)$ ?

### Answer & Explanation

Amina Hall

Beginner2022-01-22Added 11 answers

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:
$\mathrm{tan}\alpha =\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{13}\cdot \frac{13}{5}=\frac{12}{5}$

Aaron Hughes

Beginner2022-01-23Added 13 answers

$\mathrm{tan}\left(\mathrm{arccos}\left(\frac{5}{13}\right)\right)$
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: $2\frac{2}{5}$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?