Question

score on last try: 0 of 1 pts. see details for more. get a similar question you can retry this question below find the value of x. (figure is not to scale.) 13 28° x

Explanation:

Step1: Identify trigonometric ratio

We have a right triangle with adjacent side to the \(28^\circ\) angle as \(x\) and opposite side as \(13\)? Wait, no, wait. Wait, the right angle, the angle of \(28^\circ\), the side adjacent to \(28^\circ\) is \(x\)? Wait, no, let's look again. Wait, the triangle: right angle, one angle \(28^\circ\), the side opposite to \(28^\circ\) is \(13\), and the side adjacent is \(x\)? Wait, no, tangent is opposite over adjacent. Wait, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). So \(\theta = 28^\circ\), opposite side is \(13\), adjacent is \(x\)? Wait, no, maybe I got it reversed. Wait, the side labeled \(13\) is opposite the \(28^\circ\) angle? Wait, no, let's see: the right angle, the angle \(28^\circ\), so the sides: the side opposite \(28^\circ\) is \(13\), and the side adjacent is \(x\)? Wait, no, maybe it's \(\tan(28^\circ)=\frac{13}{x}\)? Wait, no, wait, maybe I mixed up. Wait, in a right triangle, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). So if the angle is \(28^\circ\), and the side opposite is \(13\), and the side adjacent is \(x\), then \(\tan(28^\circ)=\frac{13}{x}\), so \(x=\frac{13}{\tan(28^\circ)}\). Wait, no, maybe the side adjacent is \(13\) and the side opposite is \(x\)? Wait, the figure: the horizontal side (adjacent to \(28^\circ\)) is \(x\), and the vertical side (opposite) is \(13\)? Wait, no, the right angle is on the left, so the horizontal side is adjacent to \(28^\circ\), vertical is opposite? Wait, no, the angle \(28^\circ\) is at the right end, so the side adjacent to \(28^\circ\) is the horizontal side (length \(x\))? No, wait, no. Let's clarify: in a right triangle, for angle \(\theta\), \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). So if \(\theta = 28^\circ\), the opposite side is the vertical side (length \(13\)), and the adjacent side is the horizontal side (length \(x\))? Wait, no, that would be \(\tan(28^\circ)=\frac{13}{x}\), so \(x=\frac{13}{\tan(28^\circ)}\). Wait, but maybe I have it reversed. Wait, maybe the side adjacent is \(13\) and the side opposite is \(x\). Wait, let's check the labels again. The triangle: right angle on the left, so the two legs: one is vertical (length \(x\)), one is horizontal (length \(13\)), and the angle at the right is \(28^\circ\). So the angle \(28^\circ\) has adjacent side \(13\) (horizontal) and opposite side \(x\) (vertical). Ah, that makes sense! So \(\tan(28^\circ)=\frac{x}{13}\), so \(x = 13\times\tan(28^\circ)\). Wait, that's the mistake. So the angle is \(28^\circ\), adjacent side is \(13\) (horizontal), opposite side is \(x\) (vertical). So \(\tan(28^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{13}\), so \(x = 13\times\tan(28^\circ)\). Let's calculate that. \(\tan(28^\circ)\approx0.5317\), so \(x\approx13\times0.5317\approx6.9121\)? Wait, no, that can't be. Wait, maybe I got the angle's adjacent and opposite wrong. Wait, let's draw it mentally: right angle at left, so vertices: left (right angle), bottom (x), right (28 degrees). So the sides: left to bottom is vertical (x), left to right is horizontal (13), bottom to right is hypotenuse. So angle at right is 28 degrees, so the angle between the horizontal side (13) and hypotenuse is 28 degrees. So in that case, the opposite side to 28 degrees is x (vertical), adjacent is 13 (horizontal). So \(\tan(28^\circ)=\frac{x}{13}\), so \(x = 13\tan(28^\circ)\). Let's compute \(\tan(28^\circ)\). Using calculator, \(\tan(28^\circ)\approx0.5317\), so \(x\approx13\times0.5317\approx6.91\). Wait, but maybe it's \(\cot(28^\circ)=\frac{x}{13}\), which is \(\frac{1}{\tan(28^\circ)}=\frac{x}{13}\), so \(x = 13\cot(28^\circ)\). Wait, no, I think I messed up the angle's sides. Wait, let's use tangent correctly. Let's define: in triangle \(ABC\), right-angled at \(A\), angle at \(C\) is \(28^\circ\), \(AB = x\) (vertical leg), \(AC = 13\) (horizontal leg). Then angle at \(C\) is \(28^\circ\), so \(\tan(\angle C)=\frac{AB}{AC}=\frac{x}{13}\), so \(x = 13\tan(28^\circ)\). Calculating \(\tan(28^\circ)\approx0.5317\), so \(x\approx13\times0.5317\approx6.91\). Wait, but maybe the side opposite is 13 and adjacent is x, so \(\tan(28^\circ)=\frac{13}{x}\), so \(x=\frac{13}{\tan(28^\circ)}\approx\frac{13}{0.5317}\approx24.45\). Wait, now I'm confused. Let's check the figure again. The user provided a figure: right angle on the left, horizontal side (length x) at the bottom, vertical side (length 13) on the left? No, wait, the right angle is on the left, so the two legs: one vertical (left), one horizontal (bottom), and the hypotenuse connecting the top of vertical to end of horizontal. The angle at the end of the horizontal (right end) is \(28^\circ\). So the vertical leg is length 13, horizontal leg is length x. So angle at right: between hypotenuse and horizontal leg (x) is \(28^\circ\). So in that case, the opposite side to \(28^\circ\) is the vertical leg (13), and adjacent side is horizontal leg (x). Therefore, \(\tan(28^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{13}{x}\), so \(x=\frac{13}{\tan(28^\circ)}\). Now, \(\tan(28^\circ)\approx0.5317\), so \(x\approx\frac{13}{0.5317}\approx24.45\). Ah, that makes more sense. So I had the adjacent and opposite reversed earlier. So the correct formula is \(x = \frac{13}{\tan(28^\circ)}\). Let's compute that. \(\tan(28^\circ)\approx0.5317094317\), so \(13\div0.5317094317\approx24.45\). So \(x\approx24.45\) (or more precise calculation).

Step2: Calculate the value

Using a calculator, \(\tan(28^\circ)\approx0.5317\). Then \(x=\frac{13}{\tan(28^\circ)}\approx\frac{13}{0.5317}\approx24.45\).

Answer:

\(x\approx24.45\) (or more precisely, using calculator: \(\tan(28^\circ)\approx0.5317094317\), \(13\div0.5317094317\approx24.45\))