Example 2
Completion requirements
| Example 2 |
Solve for \(\theta\), given sin \(\theta = -0.819152\), where \(0^\circ \le \theta < 360^\circ\). Round to the nearest tenth of a degree.
Step 1: Determine in which quadrant(s) the solution(s) will lie.
Recall that sin \(\theta\) is negative in Quadrants III and IV.
Step 2: Determine the reference angle.
Always use the positive value of the ratio to determine the reference angle!
\(\begin{align}
\sin \theta &= 0.819152 \\
\theta &= \sin ^{ - 1} \left( {0.819152} \right) \\
\theta &= 54.999...^\circ \\
\theta &\doteq 55.0^\circ \\
\end{align}\)
Step 3: Sketch the reference angle(s) in the appropriate quadrant(s), and then determine the angle(s) in standard position.
Recall that sin \(\theta\) is negative in Quadrants III and IV.
Step 2: Determine the reference angle.
Always use the positive value of the ratio to determine the reference angle!
\(\begin{align}
\sin \theta &= 0.819152 \\
\theta &= \sin ^{ - 1} \left( {0.819152} \right) \\
\theta &= 54.999...^\circ \\
\theta &\doteq 55.0^\circ \\
\end{align}\)
Step 3: Sketch the reference angle(s) in the appropriate quadrant(s), and then determine the angle(s) in standard position.
Quadrant III:
\(\begin{align}
\theta - 180^\circ &= \theta _R \\
\theta &\doteq 180^\circ + 55.0^\circ \\
\theta &\doteq 235.0^\circ \\
\end{align}\)
Quadrant IV:
\(\begin{align}
360^\circ - \theta &= \theta _R \\
360^\circ - \theta &\doteq 55.0^\circ \\
305.0^\circ &\doteq \theta \\
\end{align}\)
Therefore, \(\theta \doteq 235.0^\circ\) and \(\theta \doteq 305.0^\circ \).
\(\begin{align}
\theta - 180^\circ &= \theta _R \\
\theta &\doteq 180^\circ + 55.0^\circ \\
\theta &\doteq 235.0^\circ \\
\end{align}\)
Quadrant IV:
\(\begin{align}
360^\circ - \theta &= \theta _R \\
360^\circ - \theta &\doteq 55.0^\circ \\
305.0^\circ &\doteq \theta \\
\end{align}\)
Therefore, \(\theta \doteq 235.0^\circ\) and \(\theta \doteq 305.0^\circ \).