Question

1. a right triangle with one angle 51 degrees, one leg length 8, and the other leg (adjacent to 51 degrees) labeled x.

Explanation:

Step1: Identify trigonometric ratio

We have a right triangle with an angle of \(51^\circ\), the opposite side to the angle is \(x\) (wait, no, wait: the adjacent side to \(51^\circ\) is \(x\)? Wait, no, the right angle, the side of length 8 is adjacent to the \(51^\circ\) angle? Wait, no, let's check: the right angle is between \(x\) and 8. So the angle \(51^\circ\) has adjacent side \(x\)? No, wait, the side opposite to \(51^\circ\) is 8? Wait, no, let's label: in a right triangle, for angle \(51^\circ\), the sides: the side adjacent to \(51^\circ\) is \(x\) (horizontal), the side opposite is 8 (vertical), and the hypotenuse is the other side. Wait, no, wait: the right angle is at the top, so the triangle has vertices: right angle at top, \(51^\circ\) at left, and the other angle at bottom. So the sides: horizontal side (top) is \(x\) (adjacent to \(51^\circ\)), vertical side (right) is 8 (opposite to \(51^\circ\)), and the hypotenuse is the left side. Wait, no, tangent of an angle in a right triangle is \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). So \(\tan(51^\circ)=\frac{8}{x}\)? Wait, no, wait: angle \(51^\circ\) is at the left, so the opposite side is the vertical side (length 8), and the adjacent side is the horizontal side (length \(x\)). Wait, no, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\), so \(\tan(51^\circ)=\frac{8}{x}\)? Wait, no, that would be if \(x\) is adjacent. Wait, no, let's correct: if the angle is \(51^\circ\), the side opposite is 8, and the side adjacent is \(x\), then \(\tan(51^\circ)=\frac{8}{x}\), so \(x = \frac{8}{\tan(51^\circ)}\). Wait, or maybe I mixed up. Wait, let's draw mentally: right angle at (0,0), (x,0), and (0,8). So the angle at (x,0) is \(51^\circ\)? No, the angle at (0,0) is \(51^\circ\). So from (0,0), the horizontal side is \(x\) (to (x,0)), vertical side is 8 (to (0,8)), and hypotenuse to (x,8). Wait, no, the right angle is at (x,8)? No, the diagram shows a right angle at the top, so the triangle has vertices: A (left, \(51^\circ\)), B (top, right angle), C (bottom, right). So AB is hypotenuse, BC is vertical (length 8), AC is horizontal (length \(x\))? No, BC is vertical (from B to C, length 8), AC is horizontal (from A to C, length \(x\)), and AB is hypotenuse. Then angle at A is \(51^\circ\), so in triangle ABC, right-angled at B, angle at A is \(51^\circ\), so \(\tan(51^\circ)=\frac{BC}{AC}=\frac{8}{x}\). Wait, no, \(\tan(\angle A)=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{AC}\), so \(\tan(51^\circ)=\frac{8}{x}\), so \(x = \frac{8}{\tan(51^\circ)}\). Wait, but let's check with calculator: \(\tan(51^\circ)\approx1.2349\), so \(x=\frac{8}{1.2349}\approx6.478\). Wait, but maybe I mixed up opposite and adjacent. Wait, if angle is \(51^\circ\), and the side adjacent is \(x\), opposite is 8, then \(\tan(51)=\frac{8}{x}\), so \(x=\frac{8}{\tan(51)}\). Alternatively, if \(x\) is opposite, then \(\tan(51)=\frac{x}{8}\), so \(x = 8\tan(51)\). Wait, that's the mistake! Let's re-express: in the right triangle, angle \(51^\circ\), the side adjacent to \(51^\circ\) is 8? No, the diagram: the right angle is at the top, so the vertical side is 8 (from top to bottom), and the horizontal side is \(x\) (from top to left). So angle at left is \(51^\circ\), so the adjacent side to \(51^\circ\) is \(x\) (horizontal), and the opposite side is 8 (vertical). Wait, no, adjacent is the side next to the angle, not the hypotenuse. So angle \(51^\circ\) has two sides: one is the hypotenuse, the other is the adjacent side (horizontal, \(x\)), and the opposite side is vertical (8). So \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{8}{x}\)? No, that would mean \(x=\frac{8}{\tan(51)}\), but let's check with \(\tan(51)\approx1.2349\), so \(x\approx6.47\). But maybe I had it reversed. Wait, maybe the angle is at the bottom, no, the diagram shows the right angle at the top, \(51^\circ\) at the left, so the sides: left angle \(51^\circ\), right angle at top, so the sides: horizontal (top) is \(x\) (adjacent to \(51^\circ\)), vertical (right) is 8 (opposite to \(51^\circ\)), hypotenuse is the left side. So \(\tan(51^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{8}{x}\), so \(x=\frac{8}{\tan(51^\circ)}\). Wait, but let's calculate \(\tan(51^\circ)\): \(\tan(51)\approx1.234897\), so \(x=\frac{8}{1.234897}\approx6.478\), which is approximately 6.48. Alternatively, if I had it reversed, \(\tan(51)=\frac{x}{8}\), then \(x=8\tan(51)\approx8\times1.2349\approx9.879\). Wait, that's different. So which is correct? Let's look at the diagram again: the right angle is at the top, so the vertical side is 8 (from top to bottom), horizontal side is \(x\) (from top to left), and the hypotenuse is from left to bottom. So the angle at the left is \(51^\circ\), so the sides: adjacent to \(51^\circ\) is \(x\) (horizontal), opposite is 8 (vertical). So \(\tan(51)=\frac{opposite}{adjacent}=\frac{8}{x}\), so \(x=\frac{8}{\tan(51)}\). Wait, but maybe the angle is at the bottom? No, the diagram shows the \(51^\circ\) at the left. So let's confirm with trigonometric ratios. In a right triangle, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). So if \(\theta=51^\circ\), opposite side is 8, adjacent is \(x\), then \(\tan(51)=\frac{8}{x}\), so \(x=\frac{8}{\tan(51)}\). Calculating that: \(\tan(51^\circ)\approx1.2349\), so \(x\approx\frac{8}{1.2349}\approx6.478\), which is approximately 6.48. Alternatively, maybe I made a mistake in identifying opposite and adjacent. Let's try \(\cot(51^\circ)=\frac{x}{8}\), since \(\cot(\theta)=\frac{\text{adjacent}}{\text{opposite}}\), so \(x = 8\cot(51^\circ)\). Since \(\cot(51^\circ)=\frac{1}{\tan(51^\circ)}\approx0.8098\), so \(x=8\times0.8098\approx6.478\), same as before. So that's correct.

Step2: Calculate \(x\)

Using \(\tan(51^\circ)=\frac{8}{x}\), we can rearrange to \(x=\frac{8}{\tan(51^\circ)}\). Using a calculator, \(\tan(51^\circ)\approx1.2349\), so \(x=\frac{8}{1.2349}\approx6.48\) (rounded to two decimal places) or more accurately, let's use more precise value of \(\tan(51^\circ)\). \(\tan(51^\circ)\) is approximately 1.2348971535, so \(x=\frac{8}{1.2348971535}\approx6.478\), which is approximately 6.48.

Answer:

\(x\approx6.48\) (or more precisely, depending on the required precision, e.g., \(x = \frac{8}{\tan(51^\circ)}\approx6.48\))