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The diagram shows a right triangle \( \triangle NPM \) with a given tangent ratio for angle \( M \), which is \( \tan M = \frac{38}{25} \). 

Here’s how to proceed:

1. **Understanding \( \tan M \):**  
   \[
   \tan M = \frac{\text{opposite}}{\text{adjacent}} = \frac{NP}{MP} = \frac{38}{25}
   \]
   Therefore, \( NP = 38 \) and \( MP = 25 \).

2. **Finding the hypotenuse \( MN \):**  
   To find \( MN \), we use the Pythagorean theorem:
   \[
   MN^2 = NP^2 + MP^2 = 38^2 + 25^2 = 1444 + 625 = 2069
   \]
   \[
   MN = \sqrt{2069} \approx 45.5
   \]

3. **Finding the angles \( \angle M \) and \( \angle N \):**  
   - \( \angle M \) can be calculated using the inverse tangent function:
     \[
     \angle M = \tan^{-1} \left( \frac{38}{25} \right) \approx \tan^{-1} (1.52) \approx 56.74^\circ
     \]
   - \( \angle N \) is complementary to \( \angle M \) since the sum of angles in a right triangle is \( 90^\circ \):
     \[
     \angle N = 90^\circ - 56.74^\circ = 33.26^\circ
     \]

Now, let's fill in the missing parts:

- \( m \angle M = 56.74^\circ \)
- \( NP = 38 \)
- \( m \angle N = 33.26^\circ \)
- \( MP = 25 \)
- \( MN \approx 45.5 \)

Do you want any more details or have any questions?

Here are 5 related questions to expand this:

1. How would the values change if \( \tan M = \frac{45}{25} \) instead?
2. Can you calculate the sine and cosine values of \( \angle M \)?
3. What if the hypotenuse \( MN \) were directly provided? How would you approach finding the other sides?
4. How does knowing one trigonometric ratio help in calculating the others?
5. How can the Pythagorean theorem be applied in non-right triangles?

**Tip:** Always remember that the sum of angles in any triangle is 180°, and in a right triangle, the two non-right angles are complementary.

The diagram shows a right triangle NPM with tan(M) = 38/25. Solve for the missing sides and angles.

Learn how to solve a right triangle with tan(M) = 38/25. This problem involves trigonometric concepts such as tangent, the Pythagorean theorem, and angle calculations using inverse trigonometry. Suitable for students in Grades 9-11.

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