Answer
$\sin \theta=\dfrac{3 \sqrt{73}}{73}$
$\cos \theta=\dfrac{8\sqrt{73}}{73}$
$\tan \theta=\dfrac{3}{8}$
$\csc \theta=\dfrac{\sqrt{73}}{3}$
$\sec \theta=\dfrac{\sqrt{73}}{8}$
$\cot \theta=\dfrac{8}{3}$
Work Step by Step
Use the Pythagorean Theorem to find the Hypotenuse. $Hypotenuse=\sqrt {3^2+8^2}=\sqrt {9+84}=\sqrt{73}$
$\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{3}{\sqrt{73}}=\dfrac{3 \sqrt{73}}{73}$
$\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{8}{\sqrt{73}}=\dfrac{8\sqrt{73}}{73}$
$\tan \theta=\dfrac{Opposite}{Adjacent}=\dfrac{3}{8}$
$\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{\sqrt{73}}{3}$
$\sec \theta=\dfrac{Hypotenuse}{Adjacent}=\dfrac{\sqrt{73}}{8}$
$\cot \theta=\dfrac{Adjacent}{Opposite}=\dfrac{8}{3}$
