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You provided the equation \(\sin \theta = -0.3090\) with \(\theta\) in the third quadrant (QIII).

In the third quadrant (QIII), the sine function is negative, which matches the given value. To find \(\theta\), we can use the inverse sine (arcsin) of the positive value to get the reference angle and then adjust for the third quadrant.

Steps:
1. Find the reference angle:
   \[
   \theta_{\text{ref}} = \arcsin(0.3090)
   \]

2. Adjust for the third quadrant:
   \[
   \theta = 180^\circ + \theta_{\text{ref}}
   \]

Let's calculate the reference angle:

\[
\theta_{\text{ref}} = \arcsin(0.3090) \approx 18^\circ
\]

Now, adding this to \(180^\circ\) to account for the third quadrant:

\[
\theta \approx 180^\circ + 18^\circ = 198^\circ
\]

Thus, \(\theta \approx 198^\circ\).

Would you like more details or have any questions about this? Here are some related questions:

1. How do you find angles in other quadrants using reference angles?
2. Can the sine function ever be positive in QIII?
3. What is the general form for angles in trigonometric functions?
4. How does the unit circle help in understanding sine values in different quadrants?
5. What is the connection between inverse sine and arc sine?

**Tip:** Remember that sine is negative in the third and fourth quadrants and positive in the first and second quadrants.

Learn how to solve the trigonometric equation sin(θ) = -0.3090 when θ is in the third quadrant (QIII). Follow the step-by-step solution using the reference angle and inverse sine function to find θ ≈ 198°. This problem is suitable for high school students (Grades 10-12) studying trigonometry.

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