But hello. I hope you're doing well. So for this problem, never graph river plain that's flying in that direction. This is our vector B, and the problem tells us that the angle between north and or the angle um, between this V vector and the Y axis the north direction here is 50 degrees instead of you vector. And then we're told that our W vector is just going straight east like that. Okay, So in order to figure out the vector components of the NW Super expressing thes vectors in terms of their X and Y components, the X component of RV vector is going to be our B times sign 50 degrees. So that's going to be the component that's in this direction the X direction. And then to get the why components of our vector v you're going to multiply the by Kasey 50 degrees. That's going to give the component of the vector B. I'm sorry. That's in this direction. The vertical direction. This is our X component right here. And this is our white component right here. So the final result for the vector of the in terms of its X and Y components would be the sign 50 degrees I plus the coast side, 50 degrees J. So that's and then R W vector would just be whatever the value of W is in the eye direction, because it's just, um, in the X direction. So now once you have that kind of would help us out with part A. So in order to find resulting vector between V and W to add two vectors together, you just add their ex components and why components together. So W is just going to be W I. So you add these two components together to get some number I. Plus, you have the J components together. In this case, there is no J component for your W vectors. You just keep this Vico signed data RV coast on 15 periods, and J. So you just add the components of vectors when you add two vectors and then in order to find the magnitude of two vectors, are the magnitude of a vector. Let's say you have a vector that's V X in the eye direction plus V Y in the J direction. Find the magnitude of the Vector V that's essentially the length of the vector going. Thio essentially do the Pythagorean theory. We're going to take the square root of the sum of the squares of the vector components, so it's going to square root V X Square plus B Y square. So that's how you get the value of the or the magnitude of up your vectors, in this case, to BB plus W. And lastly, if you're wanting to get the direction, let's say you have, um, a V X this direction b y. This direction will make this into a right triangle. And then, if you're wanting to find this angle right here, you know that the tangent of an angle theta is equal. The opposite over adjacent. So opposite in this case is V. Y. Jason's in this case is V X engine data. So that means that to find data, you would take the inverse tangent of the Y over VFW's. Okay, so let's just some review on how to solve, you know, work with factors. So now we're going to go and dive into this problem. So we're told that so again, um, and drop our figure and we have our vector V here with an angle right here. If data is equal to 50 degrees, we're told that V it's flying at a speed of 180 MPH. That's essentially the magnitude of the single, so you want to split it up into its X and Y components. Remembering what we did before. The X component of our vector here is going to be equal to our magnitude 180 times The sign of this angle sometimes sign 50 degrees that will give us the I'm component in the X direction. So now, to get the B y, the component in the Y direction, you're going to do the same thing to take your value 180 this time multiplied by the coastline of the ankle. So coastline of 50 degrees. So that means that our final answer for the vector V is going to be, uh, equal to 180 sign 50 degrees, which, if you plug that into your calculator, you'll end up with 1 37. When 89 that's in the eye direction. Then, plus your Y component will be 180 cosign 50 degrees, which, if you plug that into your calculator you get 1 15.70 that's in the J direction. So it's sort of the vector and now moving on to a W vector. That's equal Thio. So we have r W vector in this direction here. So it's just in the X direction so we don't have to even worry about this Jake opponent. We just take a magnitude of r w vector, and that's in the eye direction. So, um, are w vector. It tells us that the wind is blowing at 40 MPH. So are W vector is just 40 in the attack direction. Remember that I direction is your the same as your extraction and the J is in the Y direction. Okay, so this is our final answer for part A. This is the vector components of alright two vectors BMW that were given. So now we need to find the resulted vector V plus W. This is part a part B is V plus w So essentially add these components together. So you have V is 1. 37.89 I plus 115.7 d j w vector is gonna add that here 40 i plus and there's no J components. Zero j So if you add these together, you end up with V Plus. W is equal to adding these two I terms. Together you get 177.89 I adding these two J terms together you get plus 115.7 d j. And this is our final answer for part B of this problem. So now we need to figure out three ground speed the plane, which is just the magnitude of this V postal. So the magnitude of this view plus w it's going to be cool to the square root of the sum of the squares of these two components. Just like the Pythagorean theorem. You have a component your, um this X component of your vector in that direction. That's why component of your vector in the vertical direction. You're trying to find this high pot news value magnitude value here. So you're going to take 177 89 squared plus 115.70 squared. Take the square pretty. So you end up part C you end up with. When you plug this into your calculator, you get 212 MPH. That is the ground speed of the airplane. All right, so lastly need Thio figure out the bearing of the plains essentially the angle that it makes with north direction. So to do that, we need to take this vector here. Um, kind of draw it out. So our x component is 177.89 So 1 77 like 89 the X direction and the Y direction are white. Component is 115.7, 115.70. Okay, so we know we just got that. The magnitude of that is 212. That's its high pot news. There. We're gonna find out this angle, Fada, right here. So to do that, we know that tangent of the angle Fada is opposite over hypotenuse. Um, I'm sorry. Opposite over adjacent. So it's opposite is 115.70 over adjacent, which is 177.8. So our theta is going to be the art tan of 115 70 over 177.89 You plug that into your calculator, you get data equals 33. Um, there's just about 33 degrees. But the bearing, in order to write out the bearing of the plane, needs to figure out what the angle is with the north direction. So this is our says North. This is east. If this angle here data is equal to 33 degrees, let's say this single here, let's call it Alfa here. Is it going to be called a 90 degrees minus data. So Alfa is equal to 90 degrees minus data, which is 33 degrees, is equal to 57 degrees. That means our final answer for the bearing of this plane is from the angle with the direction of the plane of the plus W um, with the north directions 57 degrees, it's going to be north 57 degrees east. And this is our final answer right here for party. All right, so we have our answers for Parts A, B, C and D for this problem. All right, well, thanks. And I hope that helps