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