A plane is headed due south with an airspeed of 192 mph. A wind from a direction of 78.0° is blowing at 23.0 mph. Find the ground speed and resulting bearing of the plane.

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Welcome back. Everyone in this problem. We want to determine the ground speed of a fighter jet that is headed north at an air speed of 525 kilometers per hour. If the wind is blowing from a direction of 69 degrees at 70 kilometers per hour, we also want to find the resulting bearing of the jet for our answer choices. A says the resultant bearing is 335 degrees and the ground speed is 504 kilometers per hour. B says it's 350 degrees and 504 kilometers per hour respectively. C says it's seven degrees and 540 kilometers per hour respectively. And D says it's 353 degrees and 540 kilometers per hour respectively. Now, if we're going to find the ground speed of our jet and its resulting bearing, let's try to draw a diagram to visualize what's really going on here to give us some sense of direction. So since we're talking about bearings, let's draw our north and south lines. OK, there, North and Southland along with our east and west lines. And let's try to see if we can represent these speeds on our data. So we know that our fighter jet is headed north at an air speed of 525 kilometers per hour. So we can represent that speed here. Ok. And you know what, let me represent, let me put that in, in, in black because that's not the speed that we eventually want to find. We already have its ear speed. Ok. Let me make that line a bit thicker for us to see. What else do we know? Well, we know that the wind is blowing from a direction of 69 degrees at 70 kilometers per hour. So here, if we picture it, ok, where our jet is, we can also draw another set of north, south and east lines because we want to show our bearing. Ok. And at that point, let's use a, oops, let me make it a dotted line here to show that it's a secondary line because here, if we find a bearing of 69 degrees, meaning 69 degrees clockwise from the north direction, so it's probably something looking like this, then our bearing is going to look something like that. Ok. And since it's coming from 69 degrees here, then our wind will be going in this direction. Ok. And it's going at 70 kilometers per hour. So let me get rid of these here. I guess I didn't need them. Did I? The 70 kilometers per hour. Ok. Now, what else do we know? Well, we know that we're trying to find a resulting bearing off the jet and the resulting bearing would be right here. So this would be our resulting bearing. Let's call the sorry, our, our ground speed, the grown speed, the resultant vector along with the result and bearing and our result and bearing. Ok. Let me put that in red. Our result and bearing starting clockwise from the north would be all the way around here. OK. So let's first see if we can figure out the ground speed of our fighter jet and then we will try to find a result and bearing. Now, what can we know? What do we know that will help us? Well, notice here that we have a triangle formed, OK? And for our triangle, OK, we don't know any angle inside our triangle just yet. But notice here that four or 69 degree angle there is a vertical angle that corresponds to it right here. And since vertical angles are equal, that means the anger inside the triangle is also equal to 69 degrees. Let's label our triangle. So we'll have it easier to reference. So let's call the origin a, the, the, the ground speed of the fighter jet, the point that it means the wind to be and where the jet is, let's call that c Now, if we want to find our side, the ground speed of the fighter jet side A B. What do we know? We notice that we have two sides and the angle between them? That's very good because no, for a triangle, we know that we can apply the law of cosines. And the law of cosines says that the side opposite to an angle will be, or the square of the side opposite to an angle would be equal to the square of the sum of both sides minus two, multiplied by both sides multiplied by the cosine of the angle. Let's write it all, let's let's represent what we're talking about here. So for a triangle, if we label our sides as A B and see respectively, OK, then by the law of cosines by the law of cosines, we can tell that C square is going to be equal to A squared plus B squared minus two A B multiplied by the cosine of Angus C. That's the law of cosines here, we know the values of sides A B and Angus C. So we can substitute to figure out the value of side C which would be the ground speed of the fighter jet. So putting that into our formula here, we have a squared. So that's 70 squared plus B squared, 525 squared minus two multiplied by 70 multiplied by 525 multiplied by the cosine of the angle which is 69 degrees. So that will be the square of C. Therefore, that means the exact value of C then is going to be the square root of our expression, the square root of all of that 70 squared plus 525 squared minus two multiplied by 70 5, 25 plus 69. OK. So that will be the value of C. So all we need to do is to put this expression into our calculator. When you do that, what do you find? Well, you should find that the value oxide C to the nearest kilometer per hour. Because notice that's what all of our answers are written to is equal to 504 kilometers per hour. So that would be the ground speed of our jet. So this side C let's replace it here. C is 504 kilometers per hour. So we've found one part of our solution. Next, we need to find a result and bearing. Let me remove these highlights here. So again, we can look at the information that we have. So what do we know here? Well, no, we're trying to find our bearing at point A OK. And notice that if we look on our triangle, if we add the anger inside the triangle to our bearing, we'll get a complete revolution. That is, it will be 360 degrees. So if we can find the angle at a inside our triangle, then we should be able to subtract it from 360 to find the result and bearing. So how can we find this angle? Well, now notice that we know all three sides and we know an angle inside the triangle specifically, we know the angle that's opposite to 504. When we think about the angle at a, we know the side opposite to it 70 kilometers per hour. So let's think, do we know any law that uses a pair of angles and opposite side? Well, we can use the law of Science because by the law of science, the law of science basically says that the ratio of the sine of an angle to its opposite side is the same throughout the triangle. That means the ratio of the sine of 69 to 504 will be the same ratio as the sine of angle A to 70 kilometers per hour. So let's write that down and use it to solve for the sine of A because no, as we said, we're saying that the sign of A to side A is going to be equal to the s of Angus C to side C. OK? Because that's, that's what we're, that's what we're using here. So that means the sign of A 2 to 70 kilometers per hour is going to be equal to the sine of C 69 degrees to side C which is 504. That means the sine of A then is going to be equal to 70 multiplied by the sine of 69 divided by 504 which then means that here we're saying the sine of an angle equals an expression. So the angle A must be equal to the inverse sine of that expression. So A will be equal to the inverse sine of 70 s 69 degrees divided by 504. Now, we can figure that out, we can put this expression into our calculator to start for a, when we do that and round to the nearest degree, we find that A equals seven degrees. So when we go back to our diagram, then this angle here is going to be equal to seven degrees. Let me draw an arrow to it to show what it's equal to. So since that angle is equal to seven degrees, then the resultant bearing of the fighter jet is going to be equal to 360 degrees minus seven degrees. So the result and bearing is going to be equal to 360 degrees minus seven degrees, which is 353 degrees. Therefore, our fighter jet has a grown speed of 504 kilometers per hour at a resultant bearing of 353 degrees. If we look back on our answer choices, that would have been answer choice. B thanks a lot for watching everyone. I hope this video helped.

A plane is headed due south with an airspeed of 192 mph. A wind from a direction of 78.0° is blowing at 23.0 mph. Find the ground speed and resulting bearing of the plane.

Verified Solution

Key Concepts

Vector Addition

Trigonometric Functions

Bearing