Question

1. to approach a runway, a plane must begin a 7° descent starting from a height of 2 miles above the ground. to the nearest mile, how many miles from the runway is the airplane at the start of this approach?

7 mi
41 mi
16 mi
28 mi

Explanation:

Step1: Identify the trigonometric relationship

We have a right triangle where the opposite side to the \(7^\circ\) angle is the height (2 miles), and the adjacent side is the horizontal distance from the plane to the runway (let's call it \(x\)). We use the tangent function, but wait, actually, the angle given is at the top, so the angle at the bottom (the angle of depression's alternate interior angle) is also \(7^\circ\). So \(\tan(7^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{2}{x}\)? Wait, no, wait. Wait, the angle at the top is \(7^\circ\), so the angle at the bottom (the angle between the horizontal and the line of sight) is \(7^\circ\) (alternate interior angles). So the right triangle has angle \(7^\circ\), opposite side 2 miles, and adjacent side \(x\)? Wait, no, actually, the angle of depression is \(7^\circ\), so the angle inside the triangle at the plane's position (the top angle) is \(7^\circ\), so the angle at the runway end (the acute angle) is \(7^\circ\). So in the right triangle, \(\sin(7^\circ)=\frac{2}{x}\)? No, wait, let's draw it. The horizontal line is the top, the vertical line is the height (2 miles), and the hypotenuse is the distance from the plane to the runway? No, wait, the horizontal distance is \(x\), the vertical height is 2 miles, and the angle between the horizontal (dashed line) and the line of sight (the hypotenuse) is \(7^\circ\). So the angle between the hypotenuse and the vertical? No, wait, the angle given is between the dashed horizontal line and the hypotenuse, so that angle is \(7^\circ\), so the angle between the hypotenuse and the vertical is \(90^\circ - 7^\circ = 83^\circ\), but maybe it's easier to use the tangent of the angle at the bottom. Wait, the angle of depression is \(7^\circ\), so the angle of elevation from the runway to the plane is also \(7^\circ\) (alternate interior angles). So in the right triangle, the angle of elevation is \(7^\circ\), the opposite side is 2 miles (height), and the adjacent side is the horizontal distance \(x\) (the distance from the runway to the point directly below the plane). Wait, no, the question is asking for the distance from the airplane to the runway, which is the hypotenuse? Wait, no, the diagram shows a right triangle with the vertical side as 2 miles, the horizontal side as \(x\), and the angle at the top (between the horizontal dashed line and the hypotenuse) is \(7^\circ\). So the angle between the hypotenuse and the horizontal is \(7^\circ\), so the vertical side is opposite to this angle. So \(\sin(7^\circ)=\frac{2}{\text{hypotenuse}}\)? No, wait, \(\tan(7^\circ)=\frac{2}{x}\), where \(x\) is the horizontal distance, but the question is asking for the distance from the airplane to the runway, which is the length of the hypotenuse? Wait, no, the diagram's \(x\) is the hypotenuse? Wait, the diagram has the horizontal dashed line, the vertical line (height 2 miles), and the hypotenuse labeled \(x\), with the angle between the dashed line and the hypotenuse being \(7^\circ\). So in that case, the vertical side is opposite to the \(7^\circ\) angle, so \(\sin(7^\circ)=\frac{2}{x}\), so \(x = \frac{2}{\sin(7^\circ)}\). Wait, no, \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\), so if the angle is \(7^\circ\), opposite side is 2, hypotenuse is \(x\), then \(\sin(7^\circ)=\frac{2}{x}\), so \(x=\frac{2}{\sin(7^\circ)}\). But let's check with tangent. Wait, maybe I got the angle wrong. Let's re-examine. The angle between the horizontal (dashed) and the hypotenuse is \(7^\circ\), so the angle between the hypotenuse and the vertical is \(90^\circ - 7^\circ = 83^\circ\), but maybe the angle at the bottom (the acute angle) is \(7^\circ\). Wait, the right angle is at the bottom right, so the triangle has vertices: bottom left (runway), bottom right (directly below plane), top right (plane). So the angle at the top right is \(7^\circ\) (between horizontal dashed line and hypotenuse). So the angle at the bottom left (runway) is \(7^\circ\) (because the triangle's angles sum to \(180^\circ\), so \(90^\circ + 7^\circ + \text{angle at bottom left} = 180^\circ\), so angle at bottom left is \(7^\circ\)). So in that case, the opposite side to the \(7^\circ\) angle at the bottom left is the vertical side (2 miles), and the adjacent side is the horizontal side (distance from runway to directly below plane). But the question is asking for the distance from the airplane to the runway, which is the hypotenuse? Wait, no, the diagram labels \(x\) as the hypotenuse? Wait, the diagram shows the horizontal dashed line, the vertical line (height 2), and the hypotenuse \(x\) with the angle between dashed line and \(x\) being \(7^\circ\). So in that case, the vertical side is opposite to the \(7^\circ\) angle, so \(\sin(7^\circ)=\frac{2}{x}\), so \(x = \frac{2}{\sin(7^\circ)}\). But let's calculate \(\sin(7^\circ)\). \(\sin(7^\circ) \approx 0.1219\), so \(x \approx \frac{2}{0.1219} \approx 16.4\), which is about 16? Wait, no, that can't be. Wait, maybe it's \(\tan(7^\circ)=\frac{2}{x}\), where \(x\) is the horizontal distance, but the question is asking for the horizontal distance? Wait, the question says "how many miles from the runway is the airplane at the start of this approach?" So the distance from the airplane to the runway is the straight-line distance (hypotenuse) or the horizontal distance? Wait, the diagram shows \(x\) as the hypotenuse, but let's check the answer choices. The options are 7, 41, 16, 28. Let's calculate using tangent. If the angle is \(7^\circ\), the opposite side is 2, and we want the adjacent side (horizontal distance), then \(\tan(7^\circ)=\frac{2}{x}\), so \(x = \frac{2}{\tan(7^\circ)}\). \(\tan(7^\circ) \approx 0.1228\), so \(x \approx \frac{2}{0.1228} \approx 16.29\), which is about 16? But wait, that's the horizontal distance. But maybe the question is asking for the horizontal distance. Wait, let's check the diagram again. The diagram has a right triangle with right angle at the bottom right, horizontal side (adjacent) from bottom left to bottom right, vertical side (opposite) from bottom right to top right, and hypotenuse from bottom left to top right, with the angle between the horizontal dashed line (top left to top right) and the hypotenuse being \(7^\circ\). So the angle at the top right between the horizontal and hypotenuse is \(7^\circ\), so the angle at the bottom left (runway) is \(7^\circ\) (alternate interior angles, since the horizontal dashed line is parallel to the horizontal side of the triangle). So in that case, the angle at the bottom left is \(7^\circ\), opposite side is 2 (vertical), adjacent side is \(x\) (horizontal), and hypotenuse is the distance from plane to runway. Wait, no, the hypotenuse is the distance from plane to runway, so if we use \(\sin(7^\circ)=\frac{2}{\text{hypotenuse}}\), then hypotenuse \(=\frac{2}{\sin(7^\circ)} \approx \frac{2}{0.1219} \approx 16.4\), which is about 16. But wait, another way: if we use \(\tan(7^\circ)=\frac{2}{x}\), where \(x\) is the horizontal distance, then \(x \approx 16.29\), which is about 16. So the answer should be 16 mi? Wait, but let's check with \(\cot(7^\circ)\), which is \(\frac{\text{adjacent}}{\text{opposite}}=\frac{x}{2}\), so \(x = 2 \cot(7^\circ)\). \(\cot(7^\circ)=\frac{1}{\tan(7^\circ)} \approx 8.144\), so \(x = 2 \times 8.144 \approx 16.288\), which is about 16. So the answer is 16 mi.

Step1: Define the trigonometric ratio

We have a right triangle where the angle of depression is \(7^\circ\), the height (opposite side to the \(7^\circ\) angle) is 2 miles, and we need to find the horizontal distance \(x\) (adjacent side) from the runway to the point directly below the plane (or the distance from the airplane to the runway, depending on the diagram). Using the tangent function: \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\). Here, \(\theta = 7^\circ\), opposite \(= 2\) miles, adjacent \(= x\). So \(\tan(7^\circ) = \frac{2}{x}\).

Step2: Solve for \(x\)

Rearranging the formula: \(x = \frac{2}{\tan(7^\circ)}\). Calculate \(\tan(7^\circ) \approx 0.1228\). Then \(x \approx \frac{2}{0.1228} \approx 16.29\). Rounding to the nearest mile, \(x \approx 16\) miles.

Answer:

16 mi