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The image contains a trigonometry problem involving a right triangle. Here's the text extracted from the image:

**Problem Statement:**
"Use trigonometric ratios to find the angle \( \theta \) to the nearest degree. Show all calculations. Sketch any auxiliary line that you have used to support your work."

**Triangle Information:**
The triangle is a right triangle with:
- The angle \( \theta \) marked inside the triangle.
- The adjacent side to angle \( \theta \) is labeled as 100 units.
- The opposite side to angle \( \theta \) is labeled as 60 units.

**Solution Approach:**
To find the angle \( \theta \), you can use the tangent function, which is defined as:

\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]

Here, the opposite side is 60, and the adjacent side is 100. Therefore:

\[
\tan(\theta) = \frac{60}{100} = 0.6
\]

Now, to find \( \theta \):

\[
\theta = \tan^{-1}(0.6)
\]

Using a calculator to find the arctan (inverse tangent) of 0.6:

\[
\theta \approx 30.96^\circ
\]

Rounding to the nearest degree:

\[
\theta \approx 31^\circ
\]

**Final Answer:**
The angle \( \theta \) is approximately \( 31^\circ \).

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Would you like further details or explanations? 

Here are five questions related to this problem:
1. What other trigonometric ratios could you use if different sides were given?
2. How would you find the hypotenuse if it wasn't provided?
3. How does the unit circle relate to the trigonometric functions used here?
4. What is the significance of using the inverse trigonometric functions?
5. How would the calculation change if the triangle was not a right triangle?

**Tip:** Always double-check which sides (opposite, adjacent, hypotenuse) correspond to the angle you are working with, especially in non-standard triangle orientations.

Use trigonometric ratios to find the angle θ to the nearest degree. Show all calculations. Sketch any auxiliary line that you have used to support your work.

Solve a trigonometry problem by finding the angle θ in a right triangle using the tangent function. This detailed solution includes step-by-step calculations and explanations, ideal for students in Grades 9-10.

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