Find a value of theta in the interval ​[0degrees​,90degrees​] that satisfies the given statement.
tan thetaequals0.60043199


We are given that \( \tan(\theta) = 0.60043199 \) and need to find the value of \( \theta \) in the interval \([0^\circ, 90^\circ]\).

To find \( \theta \), we take the inverse tangent (arctan) of 0.60043199:

\[
\theta = \tan^{-1}(0.60043199)
\]

Let me calculate the value of \( \theta \).The value of \( \theta \) that satisfies \( \tan(\theta) = 0.60043199 \) in the interval \([0^\circ, 90^\circ]\) is approximately \( 30.98^\circ \).

Do you need more details or have any further questions?

Here are 5 related questions for further exploration:
1. What is the cosine of \( \theta \) when \( \theta = 30.98^\circ \)?
2. How would you calculate the tangent of angles greater than \( 90^\circ \)?
3. What is the inverse tangent of negative values?
4. How does the tangent function behave as \( \theta \) approaches \( 90^\circ \)?
5. What other trigonometric functions could you use to solve for \( \theta \)?

**Tip:** The tangent function is periodic with a period of \( 180^\circ \), meaning it repeats its values every \( 180^\circ \).

Find a value of theta in the interval ​[0degrees​,90degrees​] that satisfies the given statement: tan(theta) = 0.60043199.

Learn how to find the value of θ in the interval [0°, 90°] when tan(θ) = 0.60043199. This problem involves using the inverse tangent (arctan) function and is suitable for high school students studying trigonometry. The solution includes detailed steps and related questions for deeper exploration.

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