Algebra 2 (1st Edition)

$\sin \theta=\dfrac{8}{11}$ $\cos \theta=\dfrac{\sqrt{57}}{11}$ $\tan \theta=\dfrac{8 \sqrt{57}}{57}$ $\csc \theta=\dfrac{11}{8}$ $\sec \theta=\dfrac{11\sqrt{57}}{57}$ $\cot \theta=\dfrac{8 \sqrt{57}}{57}$
We will have to use the Pythagorean Theorem to find the last side. Let $a$ be the last side, then $a=\sqrt {11^2-8^2}=\sqrt {121-64}=\sqrt{57}$ $\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{8}{11}$ $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{\sqrt{57}}{11}$ $\tan \theta=\dfrac{Opposite}{Adjacent}=\dfrac{8}{\sqrt{57}}=\dfrac{8 \sqrt{57}}{57}$ $\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{11}{8}$ $\sec \theta=\dfrac{Hypotenuse}{Adjacent}=\dfrac{11\sqrt{57}}{57}$ $\cot \theta=\dfrac{Adjacent}{Opposite}=\dfrac{8 \sqrt{57}}{57}$