Evaluate -99 + 3^2•5
1 answer:
You might be interested in
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