A plane flies due south from an airport for 92 miles, and then turns and goes S63°W for 50
miles. How far is the plane from the airport (to the nearest tenth of a mile)? What direction is the plane
from the airport (to the nearest tenth of a degree)? (That is, find the bearing from the airport to the plane.)
As usual, draw a diagram. To find the distance z, it's a simple law of cosines problem.
z^2 = 92^2 + 50^2 - 2*92*cos117°
z = 123.048
Getting the bearing is a bit more work. You have to use x- and y-components of the vectors so you can add them up. Starting at (0,0), you have
<0,-90>+<50cos(-153°),50sin(-153°)>
= <-44.55,-112.7>
so, using the standard angles,
tanθ = -44.55/-112.7
θ = -112°
As a bearing from due north, that is 90-θ:
202° or S22°W
To find the distance and direction of the plane from the airport, we can use trigonometry and vector addition.
First, let's visualize the problem. The plane flies 92 miles due south from the airport, and then 50 miles in a direction S63°W. This means that the plane forms a right triangle with the airport at the right angle, where the legs of the triangle are the distances traveled in each direction.
To find the distance of the plane from the airport, we can use the Pythagorean theorem. The sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. In this case, the legs are 92 miles and 50 miles, so we have:
√(92^2 + 50^2) = √(8464 + 2500) = √10964 ≈ 104.7 miles.
Therefore, the distance of the plane from the airport is approximately 104.7 miles (to the nearest tenth of a mile).
To find the direction of the plane from the airport, we can use trigonometry. We can find the angle between the line connecting the airport and the plane, and the line parallel to the north-south direction.
The angle can be found using the inverse tangent function (tan^-1), where:
tan^-1(50/92) ≈ 28.2°.
This angle represents the bearing from the airport to the plane. However, we need to adjust it based on the given direction (S63°W). Since the given direction is measured westward from the south, we subtract the given angle from 180°:
180° - 63° = 117°.
Therefore, the direction of the plane from the airport (bearing) is approximately 117° (to the nearest tenth of a degree).
So, to summarize:
- The distance of the plane from the airport is approximately 104.7 miles (to the nearest tenth of a mile).
- The direction (bearing) of the plane from the airport is approximately 117° (to the nearest tenth of a degree).