The image contains two math problems related to right triangles. Let's solve them step by step:

Problem 5: Determine the measure of side length $x$.

The triangle has a hypotenuse of 5 m and one leg measuring 3 m. By the Pythagorean theorem:

$c^2 = a^2 + b^2$

Here, $c = 5$, $a = 3$, and $b = x$ (the side we are solving for). Substituting into the formula:

$5^2 = 3^2 + x^2$

$25 = 9 + x^2$

$x^2 = 25 - 9$

$x^2 = 16$

$x = \sqrt{16} = 4$

Answer: $\boxed{4.0 \, \text{m}}$.

Problem 6: Determine the measure of angle $\theta$.

The triangle has one leg measuring 1.0 m (opposite to $\theta$) and another leg measuring 1.5 m (adjacent to $\theta$). Using the tangent function:

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

$\tan(\theta) = \frac{1.0}{1.5}$

$\tan(\theta) = \frac{2}{3} \approx 0.6667$

To find $\theta$, take the inverse tangent ($\tan^{-1}$):

$\theta = \tan^{-1}(0.6667)$

Using a calculator:

$\theta \approx 33.69^\circ$

Rounding to the nearest degree:

Answer: $\boxed{34^\circ}$.

Let me know if you have any questions or need further explanations!

Related Questions:

1. What is the Pythagorean theorem, and when is it used?
2. How do you calculate the sine, cosine, and tangent of an angle in a right triangle?
3. How do you determine the hypotenuse if both legs are known?
4. What is the difference between radians and degrees when measuring angles?
5. Can you use trigonometric functions to solve triangles that are not right triangles?

Tip: Always double-check your triangle's labels (opposite, adjacent, hypotenuse) before applying trigonometric formulas.