### Learn math, science, English SAT & ACT from

high-quaility study
videos by expert teachers

##### Thank you for watching the preview.

To unlock all 5,300 videos, start your free trial.

# Navigation Problems - Problem 3

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let’s try a hard problem. We have a plane travelling at 540 miles per hour, at a heading of 290 degrees. The wind is 90 miles per hour from 165 degrees. Find the course and ground speed.

First of all remember that all these angles are measured clockwise from north. That’s really important in navigation problems. So let me draw that, I’ll draw my true north direction.

Now the heading is 290 degrees, that’s 20 degrees past 270. So it will look something like this. Again all directions are measured clockwise from the north, so this is going to be 290 degrees. And the magnitude of our vector is going to be the speed the air speed, 540. So I put 540 here, and then the wind is coming from 165 degrees. That’s 15 degrees short of 180, and it's 90 miles per hour so let me draw that in. Again 15 degrees short of 180 looks like this, it’s coming from that direction. Its always what they give you, the direction it's coming from. And the wind speed is 90, that’s the magnitude of this vector, so I’ll put 90 here and this direction course 165.

Now what I’m looking for in the course and ground speed is the resultant of these two vectors. So the sum of the two vectors. Let me copy this vector over here, just a rough copy. I’m going to use the head to tail method for adding these guys. So this is 90, and I want to draw the resultant. The resultant starts from here and goes in this direction.

Remember what the resultant gives you. The heading tells you what direction the airplane is travelling, how fast it's travelling with respect to the air. But the wind is blowing the plane of course a little bit. So this is going to be the result, the plane even though it's facing this way actually travels along this vector. So we want to find the magnitude of this, and we want to find its direction angle. First let’s find the magnitude, G. Now we can do that using the law of cosines if we know one more piece of information. We need to know an angle we have two sides and that’s it. So let’s try to find an angle here.

We’ve got 165 degrees and 290 degrees, so this angle in here is going to be 125 degrees. Now these two lines are parallel. Two parallel lines cut by a transversal make congruent alternate interior angles, so this is going to be 125. Now I have enough information. Let me use the law of cosines to solve this.

So G², equals 90², plus 540², minus 2 times 90 times 540, times the cosine of 125. I'll use my calculator for that. 90² plus 540² minus 2 times 90 times 540, times cosine of 125, I get 335,451.6. And then G, the ground speed will be the square root of that.

So a second square root 596.2 miles per hour. So the wind is giving the airplane a little bit of a bump, it’s got a little bit of a tail wind, and that means the ground speed is actually a little more than the air speed. Now let’s find the course. In order to find the course I’m first going to find this angle here. Now how is that going to be related to the course?

Remember the course is the direction of this vector. Well this angle 290 degrees plus theta will equal my course. So all I have to do is find theta and add that to 290. So let’s do it.

I can find theta using the law of sines. Sine of theta over 90, equals sine of 125 over G. So I multiply both sides by 90 and I get sine theta equals 90 sine 125 over G. And that tells me that theta is inverse sine of 90, sine 125 over G.

All I have to do I evaluate this is on my calculator and I still have that number G stored on my calculator. So it's inverse sine 125 divided by G. I get 7.1 degrees approximately and that means that the course, remember, before I said, the course which is this angle in red, this 290 degrees plus theta. So I just have to add this to 290 and I get approximately 297.1 degrees. That’s the course.

These problems are a little cumbersome when the angles are greater than 180 degrees, but sometimes that’s the way it works. In the end our ground speed is 596. 2 miles per hour, and our course is 297.1 degrees from north.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

##### Concept (1)

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (1)

Please Sign in or Sign up to add your comment.

## ·

Delete

## NIPPLE · 1 year, 3 months ago

Fantastic examples, thank you a bunch.