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The given problem involves a right-angled triangle with specific angle measurements. Let's break down and solve the questions.

### Problem 1:
You are given:
- \( \overline{CA} \perp \overline{AR} \), which means \( \angle 2 \) is a right angle (90°).
- \( m \angle 1 = 56^\circ \).

Since \( \angle 1 + \angle 2 \) form a right angle (because \( CA \perp AR \)), their sum must be 90°.

\[
m \angle 1 + m \angle 2 = 90^\circ
\]

Substituting the value of \( m \angle 1 = 56^\circ \):

\[
56^\circ + m \angle 2 = 90^\circ
\]

Now solve for \( m \angle 2 \):

\[
m \angle 2 = 90^\circ - 56^\circ = 34^\circ
\]

Thus, \( m \angle 2 = 34^\circ \).

### Problem 2:
You are given:
- \( m \angle 1 : m \angle 2 = 7 : 2 \), which is a ratio.
- We know \( m \angle 1 + m \angle 2 = 90^\circ \) since they are complementary angles.

Let \( m \angle 1 = 7x \) and \( m \angle 2 = 2x \).

We can set up the equation:

\[
7x + 2x = 90^\circ
\]

Solving for \( x \):

\[
9x = 90^\circ
\]
\[
x = 10^\circ
\]

Now, find \( m \angle 2 \):

\[
m \angle 2 = 2x = 2(10^\circ) = 20^\circ
\]

Thus, \( m \angle 2 = 20^\circ \).

---

Would you like any further details on the steps or explanations? Here are some related questions for practice:

1. What is the value of \( m \angle 1 \) in Problem 2?
2. How are complementary angles defined?
3. If \( m \angle 1 = 75^\circ \), what would \( m \angle 2 \) be for complementary angles?
4. In a triangle, if two angles are 45° and 45°, what type of triangle is it?
5. Can angles greater than 90° be complementary?

**Tip:** For complementary angles, the sum of the two angles always equals 90°.

1) m ∠1 = 56°, find m ∠2. 2) m ∠1 : m ∠2 = 7 : 2, find m ∠2.

Learn how to solve for complementary angles in a right-angled triangle where ∠1 = 56°. This problem involves the complementary angle theorem and solving angle ratios. Suitable for students in Grades 7-9.

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