Answer
$54.2^{o},\qquad 234.2^{o}$
Work Step by Step
reference angles:
$\left[\begin{array}{lllll}
Quadr.: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta
\end{array}\right]$
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The calculator, set to DEGREES, for
$\tan^{-1}$ ($1.3865342$) returns
54.2000000983
$\approx 54.2^{o}$
This is the reference angle, in quadrant I.
Since tangent is also positive in quadrant III,
and for $\theta$ in quadrant III, we calculate the reference angle with
$\theta^{\prime}=\theta-180^{o}$,
we find the angle $\theta$ with
$\theta=180^{\mathrm{o}}+\theta^{\prime}$
$=180^{\mathrm{o}}+54.2^{o}=234.2^{o}$
