Answer :

irspow
tanα=y/x

α=arctan(y/x), we are given the point (3.2, 6.2) so:

α=arctan(6.2/3.2)°

α=arctan(1.9375)°

α≈62.7°  (to nearest tenth of a degree)
[tex]\bf \begin{array}{rllll} (3.2&,&6.2)\\ x&&y \end{array}\\\\ -------------------------------\\\\ sin(\theta)=\cfrac{y}{r} \qquad % cosine cos(\theta)=\cfrac{x}{r} \qquad % tangent tan(\theta)=\cfrac{y}{x}\\\\ -------------------------------\\\\ tan(\theta)=\cfrac{y}{x}\implies tan^{-1}[tan(\theta )]=tan^{-1}\left( \frac{y}{x} \right) \\\\\\ \theta =tan^{-1}\left( \frac{y}{x} \right)\implies \measuredangle \theta =tan^{-1}\left( \frac{6.2}{3.2} \right)[/tex]

make sure your calculator is in Degree mode

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