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°.