blitzbabeiy

2022-01-21

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

Amina Hall

Beginner2022-01-22Added 11 answers

If you set $\alpha =\mathrm{arccos}\left(\frac{5}{13}\right)\Rightarrow \mathrm{cos}\alpha =\frac{5}{13}$ than we have to calculate:

$\mathrm{tan}\alpha =\frac{\mathrm{sin}\alpha}{\mathrm{cos}\alpha}$

$\mathrm{cos}\alpha =\frac{5}{13}\Rightarrow \mathrm{sin}\alpha =\sqrt{1-{\mathrm{cos}}^{2}\alpha}=\sqrt{1-\frac{25}{169}}$

$=\sqrt{\frac{169-25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$

So:

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

So:

Aaron Hughes

Beginner2022-01-23Added 13 answers

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