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2 answers:
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Notice the picture,
recall your SOH, CAH, TOA
you have,
opposite side, 165
adjacent side, 50
and the angle
that means, we'll need Mrs. tangent
thus
![\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}](https://tex.z-dn.net/?f=%5Cbf%20tan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%5Cimplies%20tan%28%5Ctheta%29%3D%5Ccfrac%7B165%7D%7B50%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%20tan%5E%7B-1%7D%5Cleft%5B%20tan%28%5Ctheta%29%20%5Cright%5D%3Dtan%5E%7B-1%7D%5Cleft%5B%20%5Ccfrac%7B165%7D%7B50%7D%20%5Cright%5D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%5Ctheta%3Dtan%5E%7B-1%7D%5Cleft%5B%20%5Ccfrac%7B165%7D%7B50%7D%5Cright%5D%0A%5C%5C%20%5Cuparrow%20%20%5C%5C%0A%20%5Ctextit%7Bangle%20of%20elevation%7D%5Ciff%5Ctextit%7Bangle%20of%20depression%7D)
recall your SOH, CAH, TOA
you have,
opposite side, 165
adjacent side, 50
and the angle
that means, we'll need Mrs. tangent
thus