notice the picture
we have, the opposite side
the angle
and we want the hypotenuse
so recall your SOH CAH TOA \(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\ \quad \\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent} =\)
which one has all that? low and behold, is Ms Sine,
so let's bother Ms Sine
\(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\implies sin(27^o)=\cfrac{18}{hypotenuse}
\\\\\\
hypotenuse=\cfrac{18}{sin(27^o)}\)
make sure your calculator is in Degree mode, since the angle here is in degrees, as opposed to Radian mode
