notice the picture,
recall your SOH, CAH, TOA
\(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\)
you have,
opposite side, 165
adjacent side, 50
and the angle
that means, we'll need Mrs. tangent
thusÂ
\(\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(\theta)=\cfrac{165}{50}
\\ \quad \\
tan^{-1}\left[ tan(\theta) \right]=tan^{-1}\left[ \cfrac{165}{50} \right]
\\ \quad \\
\theta=tan^{-1}\left[ \cfrac{165}{50}\right]
\\ \uparrow \\
\textit{angle of elevation}\iff\textit{angle of depression}\)
