Question

7.
11
39°
x
8.
9.

Explanation:

Step1: Identify trigonometric ratio

In a right triangle, $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$. Here, $\theta = 39^\circ$, opposite side is 11, adjacent is $x$. Wait, no: wait, the angle is $39^\circ$, the side opposite to $39^\circ$ is 11? Wait, no, the right angle, the angle given is $39^\circ$, so the side with length 11 is opposite to $39^\circ$? Wait, no, the right angle is between the side of length 11 and $x$. So the angle $39^\circ$: the side opposite is 11, adjacent is $x$. Wait, no, $\tan(39^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{11}{x}$? Wait, no, wait: if the angle is $39^\circ$, then the side opposite is 11, adjacent is $x$. So $\tan(39^\circ) = \frac{11}{x}$? Wait, no, maybe I mixed up. Wait, the right triangle: one leg is 11 (vertical), one leg is $x$ (horizontal), angle at the horizontal leg's end is $39^\circ$. So the angle $39^\circ$: the opposite side is 11, adjacent is $x$. So $\tan(39^\circ) = \frac{11}{x}$, so $x = \frac{11}{\tan(39^\circ)}$. Alternatively, maybe $\cot(39^\circ) = \frac{x}{11}$, so $x = 11 \cot(39^\circ)$ or $x = 11 \tan(51^\circ)$ (since $\cot(39^\circ) = \tan(51^\circ)$).

Wait, let's correct: in a right triangle, for angle $\theta$, $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$. So if $\theta = 39^\circ$, then the side opposite to $\theta$ is 11, and the side adjacent to $\theta$ is $x$? Wait, no, the right angle is between the two legs: one leg is 11 (let's say vertical), one leg is $x$ (horizontal), and the angle at the end of the horizontal leg (the acute angle) is $39^\circ$. So the angle $39^\circ$: the opposite side is the vertical leg (11), the adjacent side is the horizontal leg ($x$). So $\tan(39^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{11}{x}$. Therefore, solving for $x$: $x = \frac{11}{\tan(39^\circ)}$.

Step2: Calculate $\tan(39^\circ)$

Using a calculator, $\tan(39^\circ) \approx 0.8098$.

Step3: Compute $x$

$x = \frac{11}{0.8098} \approx 13.58$. Alternatively, if we use $\cot(39^\circ) = \frac{1}{\tan(39^\circ)} \approx 1.2349$, so $x = 11 \times 1.2349 \approx 13.58$.

Answer:

$x \approx 13.6$ (rounded to one decimal place) or more precise value depending on calculator.