## Precalculus (10th Edition)

$\sin\theta=\dfrac{5}{13}$ $\cos\theta=\dfrac{12}{13}$ $\tan\theta=\dfrac{5}{12}$ $\cot\theta=\dfrac{12}{5}$ $\sec\theta=\dfrac{13}{12}$ $\csc\theta=\dfrac{13}{5}$
Let's note: $h$=the hypotenuse $o$=the opposite side of angle $\theta$ $a$=the adjacent side of angle $\theta$ We are given: $o=5$ $a=12$ Determine the hypotenuse, using the Pythagorean Theorem: $h^2=o^2+a^2$ $h^2=5^2+12$ $h^2=169$ $h=\pm\sqrt{169}$ $h=\pm13$ Since $h$ is never negative, then $h=13$. Determine the $6$ trigonometric functions of angle $\theta$: $\sin\theta=\dfrac{o}{h}=\dfrac{5}{13}$ $\cos\theta=\dfrac{a}{h}=\dfrac{12}{13}$ $\tan\theta=\dfrac{o}{a}=\dfrac{5}{12}$ $\cot\theta=\dfrac{a}{o}=\dfrac{12}{5}$ $\sec\theta=\dfrac{h}{a}=\dfrac{13}{12}$ $\csc\theta=\dfrac{h}{o}=\dfrac{13}{5}$