The treasure map in FIGURE P3.41 gives the following directions to the buried treasure: 'Start at the old oak tree, walk due north for 500 paces, then due east for 100 paces. Dig.' But when you arrive, you find an angry dragon just north of the tree. To avoid the dragon, you set off along the yellow brick road at an angle east of north. After walking 300 paces you see an opening through the woods. In which direction should you walk, as an angle west of north, and how far, to reach the treasure?

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Welcome back, everyone. We're making observations about the following scenario here. You have that a lion stalks a deer grazing in front of a pine tree. Now, when close enough, it's going to charge the deer and the deer has two routes to take in order to reach a safe place. Now, the first route that they can possibly take is traveling 400 paces north and from there traveling 200 paces west. Now, the other route that they can take is that from the point of origin, they can take off at an angle that is 50 degrees west of north and then they will have to travel some distance to the safe spot at the end. Now, what I'm gonna do to complete this diagram is I'm actually going to make this into a square that we can observe all the sides with and then I'm going to label my point. So our point of origin is just gonna be zero, going in a counterclockwise direction. We have a, we have our safe space B, the top left corner is C, our midpoint for route two is D, bottom left corner is E. Now we are tasked with finding two things here. One, what is the magnitude of our vector DB? And what is going to be the angle that DB makes with the vertical axis? So this angle theta right here. Now, before we get started, I do wish to acknowledge the multiple choice answers on the left hand side of our screen, those are gonna be the values that we strive for. So without further ado let us begin. Well, for part one here, let me go ahead and change colors. What we have is that to find DB, we are gonna use Pythagorean's theorem, right. We have that the magnitude of DB is equal to the square root of the two sides of our right triangle with DB. So we are gonna have magnitude of CB squared plus B magnitude of CD squared. Now, what is CB? And what is DC? Well, let's start out with trying to find CB. We know just by observing that the magnitude of C A is equal to the magnitude of E oh Right. And what else we know is that this angle down here that do makes with de is 50 since we are told that this angle between A O and do is 50. Now, why is that important for us? Well, let's go ahead and expand this out real quick. We can see that C A is just CB plus B A and we can see that EO is just the magnitude of do times the sign of our angle 50. Now, what I'm gonna do is I'm gonna subtract the magnitude of B A from both sides. So then we get that CB, our desired magnitude is equal to do times the sign of 50 minus B A. We have all of these values. So let's go ahead and plug them in. This is gonna be equal to 300 times the sign of 50 minus 200 which gives us for CB a magnitude of 29.8. Great. Let me go ahead and change colors here and let us go ahead and now find what CD is well, similar to the first set of observations we had, we can see that the magnitude of CE is equal to the magnitude of A O. Let's expand this out a little bit. We can see that CD plus DE is equal to A O. Now, in order to isolate our desired value of CD, I'm gonna subtract de from both sides. Now, what this gives us is that CD is equal to A O minus de, we have all these values. So let's go ahead and plug them in that A O is equal to de is simply going to be do which is times the cosine of our angle 50. And what this gives us for CD is 207. changing colors. Again, here we are now ready to find what the magnitude of DB is DB is going to be the square root of CB, which is 29.8 squared plus CD, which is 207.16 squared, giving us a magnitude of 209.29. Great changing colors here. Once again, let's go ahead and move on to part two. What is going to be our angle? Theta? Well, we have a very simple formula for this. What we can say is that the tangent of our angle theta is just equal to the magnitude of CB divided by the magnitude of CD to get rid of the trigonometric function. On the left hand side, I am going to take the inverse tangent of both sides and we get that our angle theta is equal to the inverse tangent of CB over CD. This gives us the inverse tangent of 29. divided by 207.16 in which when you plug it into your calculator, it gives us an angle theta of 8.14 degrees. So now we have found the magnitude of the latter half of route two for our deer as well as the angle it makes with our vertical axis and both of these answers together. Give us a final answer of c Thank you all so much for watching. I hope this video helped we will see you all in the next one.

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Key Concepts

Vector Addition

Trigonometry

Coordinate System