A pilot is flying at 168 mph. She wants her flight path to be ...

Question

A pilot is flying at 168 mph. She wants her flight path to be on a bearing of 57° 40′. A wind is blowing from the south at 27.1 mph. Find the bearing she should fly, and find the plane's ground speed.

Accepted Answer

Welcome back. Everyone in this problem. Sam is flying an aerobatic plane on a windy day. The wind speed is 30 MPH and blowing from the south. The plane is flying at a speed of 212 MPH and some wants its flying path to be on a bearing of 61 degrees. 20 minutes, determine the ground speed of the plane and the bearing. He should fly to stay on that path for our answer choices. A says the bearing is 68 degrees 28 minutes and the ground speed is 225 MPH. B the bearing is 68 degrees 28 minutes and the ground speed 252 MPH. C bearing of 54 degrees 12 minutes. And the ground speed 225 MPH and d bearing 54 degrees 12 minutes and the ground speed of 252 MPH. Now, before we start to figure out the ground speed and the bearing, let's try to make sense of this problem. So we're talking about a plane flying based on particular bearings. So since we're dealing with bearings. Let's draw up our, uh, north and south and east and west lines. Ok. So, you know, this is what we're talking about. Let me make some space. Now. What do we know? Ok. Well, we know that we have a plane flying at a speed of 212 MPH. We don't know what bearing it's flying on yet. Ok. But on our diagram here showing our speeds, we know there's a wind speed of 30 MPH blowing from the south. So if we imagine our plane here, there is a wind speed going 30 MPH from the south. So it must be going directly north. Now, as we said, we don't know what bearing the plane is flying on, but we know that some wants its flying path to be on a bearing of 61 degrees 20 minutes. In other words, if we here represent it in red, let me put it in red. Ok. This is the bearing that Sam wants the plane to be flying on where it's, remember that a bearing is measured clockwise from the north. So it's 61 degrees and 20 minutes clockwise from the north. Ok. So that's what we know so far. Now, what we're trying to find is the bearing he should fly to stay on that path. In other words, we're trying to find all of this entire angle here because when we think about it, if he stays on that bearing, the north wind will push him up to the bearing that he wants to be at. And we also want to find the ground speed of the plane. Let me keep this yellow here. We want to find the ground speed of the plane, which is our red wind speed. Our red vector here representing uh the speed, OK? Or the, the red arrow, sorry, representing the speed. So let's imagine here that this represents a triangle. OK. You notice here that it does represent a triangle, a triangle with these points A B and C. Now, if we look at this triangle here, if we can figure out this unknown angle, let's call this anger theta, then we can add it to our bearing of 61 degrees 20 minutes to figure out the bearing that the plane should stay on to stay on that path. And if we figure out that angle and any other angles in our triangle, then we will be able to find the length of A B that is the ground speed of the plane. So how can we figure that out? Let's think about it here. Is there anything that we notice? Well, let's think about our triangle that we formed. Do we know anything else apart from what is labeled here? Well, notice that if I were to highlight these side here, we have the wind speed going north and we have our north line. OK. And that means we can find a pair of correspond of, of alternate angles rather because notice that here, the alternate angle to 61 degrees, 20 minutes would be Anger B. Therefore angle B is 61 degrees and 20 minutes. Now that we know the value of Anger B, could we use that to help us figure out our anger theta here. Yeah. Yes, we could notice that in our triangle. OK. Let me take off our blue highlight here in our triangle, we have a pair of opposite sides and angle. OK? And for our triangle, ABC if we label the side opposite to anger, A as A, OK? LA decide opposite to anger, B as B and labor decide opposite to anger C. That is the grown speed as C come on C then we can apply the sign rule because by the sign rule or by the law of sign, OK? By the law of sign, some people call it the sign rule, the law of science. It's basically the same thing. So by the law of science, OK? By the law of science, the law of science tells us that the ratio of an anger to its opposite side is the same through the ratio of the s of an anger rather to its opposite side is the same throughout the triangle. Now here we know that 61 degrees, 20 minutes is opposite to 212 MPH. So we can figure out the value of anger. A let's not call it theta, let's call it A here. We can figure out the value of A, ok. The value of anger is the value of an air because it is opposite to our 30 mile per hour wind speed. So by the law of science, that means the sine of a rat of the sine of A to its opposite side, 30 MPH is going to be equal to the sine of 61 degrees, 20 minutes. That is B to its opposite side, 212 MPH. Therefore, that means the sine of A is going to be equal to 30 multiplied by the sine of 61 degrees, 20 minutes all divided by 212. And if we want to solve for a, then we can use the inverse sine because no, that tells us that A is going to be equal to the inverse sine of our expression. So that's 30 s 61 degrees 20 minutes divided by 212. Now, let's put this expression in our calculator. What do you find? Well, you should find that to the nearest minute, which is here to our nearest minute. A is approximately equal to seven degrees and eight minutes. OK. So it is approximately equal to seven degrees and eight minutes. Now, what does that mean for us? Well, no, that means anger air seven degrees. Uh Let me, it's a bit small there. So seven degrees eight minutes. So now the entire angle, the bearing that some should stay on to fly, the path is going to be equal to 61 degrees 20 minutes plus seven degrees eight minutes. So therefore, the bearing stamp must stay on bearing at which some should fly is going to be equal to 61 degrees, 20 minutes plus seven degrees eight minutes, which is equal to 68 degrees and 28 minutes. Therefore, some should stay, stay on a bearing of 68 degrees and 28 minutes to stay the path. So now we have that bearing, since we have that bearing, all we need to do now is to figure out the ground speed, which in this case is side C. So now we know the angle, OK? Now we know two angles out of the three. Can we figure out the third angle here? Well, what if we knew the value of Anglesey, if we know the value of the angle at sea? OK, then we could use the cosine rule because here notice that we have two sides, an angle between them. Let me put that in red and we want to find a side that is opposite to that angle. So we can use the cosine route, the law of cosines. But before we do that, we'll need to figure out the angle. Well, we already know the other two angles in the triangle and angles in a triangle sum to 180 degrees. So no, that means Anger C OK. Yeah, I should write this. Let me get rid of the south side of it here. Let me make some space because no Anger Sea is going to be equal to 180 degrees minus the sum of the other two angles in the triangle. 61 degrees, 20 minutes plus seven degrees, eight minutes. Now, when we subtract that from 180 degrees, we get 101 degrees and 32 minutes. So angle C is 101 degrees 32 minutes. No, that's very good because here, that means by the cosine law. So by the law of cosines, OK, let me make a line here by the law. Of course, science that means the side opposite to angle CC squared is going to be equal to the square of the other two sides 212 squared. Well, let me write their values first. So C squared will be B squared plus A squared minus two A B multiplied by the cosine of angle C. It's like the Pythagorean theorem, but with a little extra in between. So that means then C squared will be 212 squared plus 30 squared minus two, multiplied by 212 multiplied by 30 multiplied by the cosine of 101 degrees and 32 minutes, OK. Now, if we were to put this into our calculator here to solve for side C that means C is going to be equal to the square root of the result of our expression here. OK. And when we put this expression into our calculator, we should get 50,000 512.78009. OK. So now we need to find the square root of that, of that value. And the square root of it is equal to 224.7504841 or, or when we round it to the nearest uh mile per hour, then we get side C to be approximately equal to 225 MPH. Ok. Therefore, the ground speed of the plane, ok. That side of the triangle that we formed is equal to 225 MPH. And the bearing at which some should fly to maintain the path is 68 degrees 28 minutes. When we look back on our answer choices, the correct choice is a thanks a lot for watching everyone. I hope this video helped.

A pilot is flying at 168 mph. She wants her flight path to be on a bearing of 57° 40′. A wind is blowing from the south at 27.1 mph. Find the bearing she should fly, and find the plane's ground speed.

Key Concepts

Vector Addition in Navigation

Bearing and Angle Measurement

Trigonometric Resolution of Vectors

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