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Determine the angle(s) in standard position that satisfy the equations.

  1. \(\sin \theta = \frac{{\sqrt 3 }}{2},\thinspace \rm {where} \thinspace 0^\circ \le \theta < 360^\circ \)

    The quadrants where \(\sin \theta\) is positive are Quadrants I and II.

    The \(\sin \theta = \frac{\sqrt{3}}{2}\) indicates that \(\theta\) is one of the special triangle angles, specifically \(\theta = 60^\circ\).  This is the reference angle.

    Quadrant I:

    \(\begin{align}
     \theta &= \theta _R  \\
     \theta &= 60^\circ  \\
     \end{align}\)

    Quadrant II:

    \(\begin{align}
     180^\circ - \theta &= \theta _R  \\
     180^\circ - \theta &= 60^\circ  \\
     \theta &= 120^\circ  \\
     \end{align}\)


  2. \(\tan \theta = -1.19175,\thinspace \rm {where} \thinspace 180^\circ \le \theta < 360^\circ \)

    The quadrant where \(\tan \theta\) is negative, from \(180^\circ\) to \(360^\circ\), is Quadrant IV.

    The reference angle is found by taking the inverse tangent of the positive value of the ratio.

    \(\begin{align}
     \tan \theta _R &= 1.19175 \\
     \theta _R &= \tan ^{ - 1} \left( {1.19175} \right) \\
     \theta _R &= 49.999...^\circ  \\
     \theta _R &\doteq 50.0^\circ  \\
     \end{align}\)

    Quadrant IV:

    \(\begin{align}
     360^\circ - \theta &= \theta _R  \\
     360^\circ - \theta &\doteq 50.0^\circ  \\
     \theta &\doteq 310.0^\circ  \\
     \end{align}\)