A luxury liner leaves port on a bearing of 110.0° and travels ...

Question

A luxury liner leaves port on a bearing of 110.0° and travels 8.8 mi. It then turns due west and travels 2.4 mi. How far is the liner from port, and what is its bearing from port?

Accepted Answer

Hello, today we are going to be solving the following problem. So we are told that a motor boat departs 10.4 miles out of the port with a bearing of 313 degrees. After a few hours, the motor makes a turn and starts moving to the west for 6.9 miles. We want to find the bearing of the ship and its distance from the port. Let's go ahead and draw an image to help us solve this problem. So the first thing that the problem tells us is that the motor board departs from the origin 10.4 miles out with a bearing of 313 degrees. So the motor boat is traveling in this direction with a bearing of 313 degrees. Then after a few hours, the motor boat makes a turn and starts moving to the west for 6.9 miles. So at some point, the boat starts heading west for a total of 6.9 miles. We want to find the bearing which is the angle of the ship and its distance from the port, which is the origin. Now, we can label this side with the vector pointing to the very end of the ship. So what we've crafted here is a roughly drawn triangle. We can label this side A, we can label this side B and we can label this side C. And one thing I forgot to label here is that A has a length of 10.4 miles. So how can we go about solving for the length of C and the bearing angle to sea? Well, we can use a little bit of right angle trigonometry. See what we can do is we can craft a right angle going from the end point of the 10.4 mile mark to the Y axis. We'll go ahead and label this side D and label this side E and since we're seeing that this is a right triangle, we're going to label the right angle located between D and E. Now, what we want to do is we want to find an angle between B and A. The reason for this be is because we can use the law of cosines to help us solve for the length C. But we'll need to find the angle between B and A first. So we're going to use this right triangle to help us find the angle between B and A. Now, what we are going to do is we're going to find the angle located just below the right angle. We know that the bearing from the Y axis to A is 313 degrees. If we can continue to the Y axis, this will be a total angle of 360 degrees. So if we do 360 minus 313 that will give us the angle with respect to the side D so 360 minus 313 will give us the angle of 47 degrees. So we know that the angle with respect to D is 47 degrees. Now we can solve for the angle with the res with respect to the side E by using the angle, some property recall that the sum of all three angles within a triangle must add to 180 degrees. So the angle E will equal to 180 degrees minus 90 degrees minus 47 degrees. After simplifying, we will have an angle of 43 degrees. So we know that the angle with respect to E is going to be 43 degrees. Now, you may be asking yourself how is this angle going to help us find the angle between B and A? Well, what we can do is we can label the sides B and D as a vertical line. Recall that the, that the angle for a vertical line is equal to 180 degrees. So if we do 180 minus 43 that will give us the angle between B and A with respect to the site C. So what we're going to do is we're going to label the angle C and set it equal to 180 degrees minus 43 degrees. 180 minus 43 will give us the angle of 137 degrees. So now we know that the angle with respect to the sight C is going to be 137 degrees. Now recall that we know the length of A to be 10.4 miles, we know the length of B to be 6.9 miles. And we know that the angle with respect to C is 137 degrees. Now, what we can do is we can use the law of cosines to help us find the length of sea. Now, the law of cosines will allow us to write that the length of C squared is equal to the side A squared plus the side B squared minus two, multiplied by the side A multiplied by the side B multiplied by cosine of angle C. We know what A is, we know what B is and we just solve for angle C. So now what we need to do is we need to plug in the values before plugging in the values. Let's take the square root of both sides of the equation that will give us site C is equal to the square root of A squared plus B squared minus two multiplied by a multiplied by C multiplied by the cosine of angle C plugging in the values will leave us with the square root of 10.4 squared plus 6.9 squared minus two, multiplied by 10.4 multiplied by 6.9 multiplied by cosine of the angle C which is 137 degrees at this point. I would recommend plugging this value into a calculator. And when doing so, you should get that, the site C is going to equal to 16.1 miles. So what this tells us is that the motor boat had traveled a total of 16.1 miles. So that's one part of the information we are looking for. Now, what we want to do is we want to find the bearing angle to this distance. So in order to find the bearing angle to this distance, we will first need to find the angle with respect to the side B, if we find this angle, we can subtract that degrees from 313 degrees. And that will give us the bearing angle to the side sea. So let's go ahead and start by solving for the angle B. In order to solve for the angle B, we can use the law of signs. The law of signs gives us three equivalent ratios. We have sign of A divided by the si length A is equal to sine of angle B divided by the S length B is equal to sine of angle C divided by the S length C. What we are going to do is we are going to pick any two of the three ratios, set them equally to each other. And use that to solve for the angle B. Now, we know that the Anglesey is 137 degrees and we just sold for the Cl Sea which is 16.1. We don't know the angle B but we do know that the S length B is equal to 6.9. With that being said, let's pick the ratios with respect to B and C set them equal to each other and use that to solve for the angle B. So doing this will allow us to write sine of B divided by SI length B is equal to sine of C divided by the Silent Sea. We can solve for the angle B by first cross multiplying the ratios doing so will allow us to write the S length C multiplied by sine of angle B is equal to the SI length B multiplied by sine of angle C. We will divide both sides of this equation by the side length C. And that will leave us with sign of angle B is equal to the S length B multiplied by sine of angle C divided by the Silent Sea. In order to solve for the angle B, we'll have to remove it from the sine equation. And in order to do that, we'll take the sine inverse of both sides of this equation. That will leave us with the angle B is equal to the sine inverse of the side length B multiplied by sine of angle C divided by the Silent Sea. Now, we will plug in the values in order to solve for the angle B. So the angle B will equal to the sine inverse of the si length B which is equal to 6.9 multiplied by sign of Anglesey, which is 137 degrees divided by the side lengthy, which we solved earlier to be 16.1. Now, I would recommend plugging this value into a calculator and when doing so, you should get an approximate angle of 17 degrees. Now, this is going to be the angle B. Now, just like before we're going to take the bearing angle of 313 degrees and subtract 17 degrees from this value doing so will give us the bearing angle for the distance of C. So we're going to take 313 degrees minus 17 degrees and that is going to give us an angle of 296 degrees. This is going to be the bearing angle for the distance that the motor boat had traveled. So just to recap, we had used the law of signs to solve for the distance that the motor boat had traveled, using the law of cosines gave us 16.1 miles. And we use the law of signs to solve for the angle B and subtracting the angle B from the original bearing angle gave us 296 degrees, which is going to be the bearing angle for the site C. And with that being said, the answer to this problem is going to be B so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

A luxury liner leaves port on a bearing of 110.0° and travels 8.8 mi. It then turns due west and travels 2.4 mi. How far is the liner from port, and what is its bearing from port?

Key Concepts

Bearing and Direction in Navigation

Vector Addition and Components

Trigonometric Functions and Resultant Magnitude

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