Solve The Problem.An Airplane Flies At A Bearing Of 10° For 180 Miles, And Then At A Bearing Of 315° For 95 Miles. How Far Is The Plane From The Starting Point, And In What Direction?A. 247 Mi, 307°B. 275 Mi, 307°C. 247 Mi, 352°D. 275 Mi, 2°

Mar 10, 2025 by ADMIN 242 views

Solving the Problem: Finding the Distance and Direction of an Airplane from its Starting Point

Introduction

In this problem, we are given the bearing and distance traveled by an airplane in two different directions. We need to find the distance of the plane from its starting point and the direction in which it is located. This problem involves the use of trigonometry and the concept of vectors.

Understanding the Problem

The airplane flies at a bearing of 10° for 180 miles, which means it travels 180 miles in a direction 10° north of east. Then, it flies at a bearing of 315° for 95 miles, which means it travels 95 miles in a direction 45° south of west.

Breaking Down the Problem

To solve this problem, we can break it down into two parts:

Finding the position of the airplane after it travels 180 miles at a bearing of 10°.
Finding the position of the airplane after it travels 95 miles at a bearing of 315°.
Using the two positions to find the distance and direction of the airplane from its starting point.

Step 1: Finding the Position of the Airplane after 180 Miles at a Bearing of 10°

To find the position of the airplane after it travels 180 miles at a bearing of 10°, we can use the concept of vectors. We can represent the direction of the airplane as a vector with a magnitude of 180 miles and an angle of 10°.

x = 180 \* \cos(10°)
y = 180 \* \sin(10°)

Step 2: Finding the Position of the Airplane after 95 Miles at a Bearing of 315°

To find the position of the airplane after it travels 95 miles at a bearing of 315°, we can use the concept of vectors. We can represent the direction of the airplane as a vector with a magnitude of 95 miles and an angle of 315°.

x = 95 \* \cos(315°)
y = 95 \* \sin(315°)

Step 3: Finding the Distance and Direction of the Airplane from its Starting Point

To find the distance and direction of the airplane from its starting point, we can use the concept of vectors. We can add the two vectors representing the position of the airplane after each leg of the flight to find the final position.

x_final = x1 + x2
y_final = y1 + y2

Step 4: Calculating the Distance and Direction

To calculate the distance and direction of the airplane from its starting point, we can use the Pythagorean theorem and the concept of trigonometry.

distance = \sqrt{x_final^2 + y_final^2}
direction = \tan^{-1}\left(\frac{y_final}{x_final}\right)

Calculating the Values

Now, let's calculate the values of x_final, y_final, distance, and direction.

x1 = 180 \* \cos(10°) = 175.92
y1 = 180 \* \sin(10°) = 30.18
x2 = 95 * \cos(315°) = -48.55
y2 = 95 * \sin(315°) = 84.55
x_final = x1 + x2 = 127.37
y_final = y1 + y2 = 114.73
distance = \sqrt{x_final^2 + y_final^2} = 155.93
direction = \tan^{-1}\left(\frac{y_final}{x_final}\right) = 34.93°

Conclusion

The distance of the airplane from its starting point is approximately 155.93 miles, and the direction is approximately 34.93° north of east.

However, this is not among the given options. Let's recalculate the values.

x1 = 180 \* \cos(10°) = 175.92
y1 = 180 \* \sin(10°) = 30.18
x2 = 95 * \cos(315°) = -48.55
y2 = 95 * \sin(315°) = 84.55
x_final = x1 + x2 = 127.37
y_final = y1 + y2 = 114.73
distance = \sqrt{x_final^2 + y_final^2} = 155.93
direction = \tan^{-1}\left(\frac{y_final}{x_final}\right) = 34.93°