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Chapter 10: Problem 2

An aeroplane flies 150 miles on a bearing of \(105^{\circ}\) and then 107 miles on a bearing of \(217^{\circ}\). Find the bearing that the aeroplane must take to fly directly back to the starting position.

Short Answer

Expert verified
Answer: To find the bearing angle, follow these steps: 1. Understand the problem. 2. Represent the information as vectors. 3. Calculate the vectors. 4. Find the required vector. 5. Calculate the coordinates of the vector. 6. Find the bearing angle using the arctangent formula and the coordinates of the required vector.

Step by step solution

01

Understanding the Problem

We are given that the aeroplane flies 150 miles on a bearing of \(105^{\circ}\) and then 107 miles on a bearing of \(217^{\circ}\). We need to find the bearing that the aeroplane must take to fly directly back to the starting position.
02

Representing the Information as Vectors

We can represent the initial 150 miles on a bearing of \(105^{\circ}\) as a vector in the plane as follows. Let A be the starting point, \(B_1\) be the point after flying 150 miles towards \(105^\circ\) and \(B_2\) be the final point after flying 107 miles towards \(217^\circ\). We need to find the bearing when flying back from the point \(B_2\) to A.
03

Calculating the Vectors

The vector from A to \(B_1\) can be represented as: \(\vec{AB_1} = 150\begin{bmatrix}\cos(105^\circ)\\\sin(105^\circ)\end{bmatrix}\). The vector from \(B_1\) to \(B_2\) can be represented as: \(\vec{B_1B_2} = 107\begin{bmatrix}\cos(217^\circ)\\\sin(217^\circ)\end{bmatrix}\). Adding the two vectors, we get the resultant vector from A to \(B_2\): \(\vec{AB_2} = \vec{AB_1} + \vec{B_1B_2}\).
04

Finding the Required Vector

To fly back directly to the starting position from \(B_2\), the required vector will be the negative of the resultant vector from A to \(B_2\): \(\vec{B_2A} = -\vec{AB_2}\).
05

Calculating the Coordinates of the Vector

Now calculate the coordinates of the vector \(\vec{AB_2}\) using the previously calculated vectors as: \(\vec{AB_2} = 150\begin{bmatrix}\cos(105^\circ)\\\sin(105^\circ)\end{bmatrix} + 107\begin{bmatrix}\cos(217^\circ)\\\sin(217^\circ)\end{bmatrix}\). Compute the result to obtain the vector \(\vec{AB_2}\).
06

Finding the Bearing Angle

Finally, we need to find the angle that represents the required bearing. To do this, use the arctangent to find the angle between the horizontal axis and the vector \(\vec{B_2A}\): \(\theta = \arctan{\frac{y}{x}}\), where \(x\) and \(y\) are the coordinates of the vector \(\vec{B_2A}\). Let's not forget to add 360 degrees or subtract 360 degrees if necessary to make the anglepositive and to stay between 0 and 360 degrees, i.e., in the first or fourth quadrant. Now we have the angle \(\theta\), which represents the bearing that the aeroplane must take to fly directly back to the starting position.

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Most popular questions from this chapter

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