## Algebra 2 (1st Edition)

$\dfrac{11 \sqrt 2}{12}$
We need to use the Pythagorean Theorem in order to solve $y$ such that $r^2=x^2+y^2 \implies y=\sqrt {r^2-x^2}$ $y=\sqrt{(11)^2-(7)^2}=\sqrt{72}=6\sqrt 2$ Since, $\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{r}{y}$ Thus, $\csc \theta=\dfrac{11}{6\sqrt{2}}=\dfrac{11 \sqrt 2}{12}$