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The image contains a math question asking for the range of values for \(\theta\) that satisfy \(\sin \theta = 0.5736\). The given answer choices are:

- A) \(30^\circ\) and \(150^\circ\)
- B) \(35^\circ\) and \(145^\circ\)
- C) \(90^\circ\)
- D) \(0^\circ\) and \(180^\circ\)

Let's solve this equation step by step.

The equation \(\sin \theta = 0.5736\) can be solved by taking the inverse sine (\(\sin^{-1}\)) of 0.5736 to find the reference angle.

We know that \(\sin \theta\) is positive in the first and second quadrants, so we'll find two solutions within the range \(0^\circ \leq \theta \leq 180^\circ\).

Now, let's calculate the angles.The solutions to \(\sin \theta = 0.5736\) are approximately:

- \( \theta_1 = 35^\circ \) (in the first quadrant)
- \( \theta_2 = 145^\circ \) (in the second quadrant)

Thus, the correct answer is **B) 35° and 145°**.

Would you like more details or have any other questions?

Here are 5 related questions:
1. How do we determine which quadrants the sine function is positive in?
2. What is the general solution for \(\sin \theta = k\) where \( -1 \leq k \leq 1 \)?
3. Why is it important to know the reference angle in trigonometry?
4. How can we find the angle if \(\sin \theta\) is negative?
5. How do we apply this process for angles in radians?

**Tip**: Always remember that the sine function is positive in both the first and second quadrants, but negative in the third and fourth quadrants.

What is the range of values for θ such that sin θ = 0.5736?

Learn how to solve the trigonometric equation sin θ = 0.5736 with a detailed step-by-step solution. This problem explores the inverse sine function and reference angles, with solutions for θ in both the first and second quadrants. Suitable for students in Grades 9-11.

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