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To solve the given right triangle, we’ll need to find:

1. \( m \angle C \) (the measure of angle \( C \)),
2. \( AB \) (the hypotenuse), and
3. \( BC \) (the adjacent side to angle \( A \)).

Given:
- \( \angle A = 62^\circ \),
- \( AC = 10 \) (opposite to \( \angle A \)),
- \( \angle B = 90^\circ \).

### Step 1: Find \( m \angle C \)
Since \( \angle A + \angle B + \angle C = 180^\circ \) and \( \angle B = 90^\circ \):
\[
\angle C = 180^\circ - 90^\circ - 62^\circ = 28^\circ
\]

Thus,
\[
m \angle C \approx 28^\circ
\]

### Step 2: Find \( AB \) (Hypotenuse)
Using the sine function with angle \( A \):
\[
\sin(62^\circ) = \frac{AC}{AB}
\]
Rearrange to solve for \( AB \):
\[
AB = \frac{AC}{\sin(62^\circ)} = \frac{10}{\sin(62^\circ)}
\]
Using \( \sin(62^\circ) \approx 0.8829 \):
\[
AB \approx \frac{10}{0.8829} \approx 11.3
\]

### Step 3: Find \( BC \) (Adjacent side to \( \angle A \))
Using the tangent function with angle \( A \):
\[
\tan(62^\circ) = \frac{AC}{BC}
\]
Rearrange to solve for \( BC \):
\[
BC = \frac{AC}{\tan(62^\circ)} = \frac{10}{\tan(62^\circ)}
\]
Using \( \tan(62^\circ) \approx 1.8807 \):
\[
BC \approx \frac{10}{1.8807} \approx 5.3
\]

### Summary of the Results
- \( m \angle C \approx 28^\circ \)
- \( AB \approx 11.3 \)
- \( BC \approx 5.3 \)

Would you like further details or have any questions about these steps?

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Here are 5 related questions to deepen understanding:

1. How would you calculate the length of \( AC \) if only \( AB \) and \( \angle A \) were given?
2. Can you use the Pythagorean theorem to verify the results for \( AB \) and \( BC \)?
3. What trigonometric function would you use to find \( BC \) if \( \angle C \) and \( AB \) were given?
4. How does rounding affect the accuracy of these results?
5. What is the relationship between the side lengths and angles in a right triangle?

**Tip:** Remember that trigonometric functions are very useful in solving right triangles when one angle and one side are known.

Solve the right triangle. Round side and angle measures to the nearest tenth.

Learn how to solve a right triangle given a 62° angle and an opposite side length of 10. Use trigonometric ratios to find the hypotenuse, adjacent side, and remaining angle. Suitable for Grades 9-10.

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