Solve each problem. See Examples 5 and 6.

Question

Bearing and Ground Speed of a Plane An airline route from San Francisco to Honolulu is on a bearing of 233.0°. A jet flying at 450 mph on that bearing encounters a wind blowing at 39.0 mph from a direction of 114.0°. Find the resulting bearing and ground speed of the plane.

Accepted Answer

Hey, everyone in this problem, we're asked to determine the ground speed of a charter plane heading from New York to L A on a bearing of 257 degrees. The air speed of the charter plane is 550 MPH. It encounters wind current from the direction of 125 degrees at 50 MPH. We're also asked to determine the resultant bearing of the plane as well. We have four answer choices A through D each containing different values for the bearing and the ground speed. And we're gonna come back to these when we are done with this problem. So to start out, let's draw out a picture of what we have going on. So we can imagine drawing our coordinate axis, ok. And we can think about the plane flying at a bearing of 257 degrees, ok. So that's gonna be measured from the positive Y axis or the northward direction, clockwise all the way around until 257 degrees. So that's the first thing we're drawing here. Then we have the air speed of 550 MPH. And so the speed we're traveling along this line is 550 MPH. Ok. Next, we have wind current, the wind current is gonna be traveling in the direction from 125 degrees. And so we go ahead and again, draw our coordinate axes. We're gonna draw it at the end, the tip of our plane so that we can add these vectors together And we have the plane and then we have this wind current. It's gonna be at 100 and 25 degrees, but it's coming from 125 degrees. So we can imagine 125 degrees. It's gonna be somewhere around here. The wind is coming from that direction. So it's gonna be traveling up into the left, it's traveling at 50 MPH. And so the actual direction that this plane is traveling in terms of the ground speed, the ground bearing is gonna be tip to tail of these vectors. We can call the ground speed VG, that's the speed we're gonna be looking for. And then the bearing we're gonna be looking for is going to be in red here and we can call that beta B. OK. So what do we want to do? First? We need to figure out some more information in this diagram before we can get started right now, we have two side lengths in the triangle that we've created. But we have no angles. Now, we can do a little bit of geometry to fix that. If you imagine this triangle in the bottom, right. OK. We form this 250 to 7 degree angle. OK. The 1st 180 degrees are taken up by forming that straight line from positive y axis to negative y axis. So if we subtract 180 degrees, we're left with 77 degrees. So that means that the angle between the vertical here and this triangle is 77 degrees if this is 77 degrees, and we have a right angle triangle. So the other angle that's left is gonna make 13 degrees. And we already know the angle that forms 77 degrees with 13 degrees is 77 degrees. And we have this full 90 degree angle, the angle from the vertical moving left towards that wind current, we know that this is gonna form a 55 degree angle because that 55 degrees and the 125 degree bearing must form that straight line 180 degrees. And so the total angle that we have inside of our triangle is gonna be that 55 degrees plus the 77 degrees. So it is gonna be will dry in green 132 degrees. So now we have two side links and the enclosed angle that's enough to use a law of cosines to find the other side links. So that's exactly what we are gonna do. So if we wanna calculate VG may recall that law of cosines C squared is equal to A squared plus B squared minus two A B cosine angle C. In this case, side C is VG that ground speed we wanna find we have VG squared is gonna be equal to one of the side links squared. So we'll say 50 MPH squared plus the other side link squared, 550 MPH squared minus two, multiplied by the first deadline, 50 MPH multiplied by the second sideline, 550 MPH multiplied by cosine of the angle between them, which we found was 132 degrees. So if we work all of this out on the right hand side and we take the square root, what we're gonna get is that that ground speed VG is gonna be 500 in 84.6385 MPH. Ok. So we're done one part of this question. We found that ground speed of the charter plane. Now, we can take a look at the answer choices. Option A B and D all have this ground speed of 585 MPH and that matches with what we found. But option C has 558 which is not what we found. So we can go ahead and eliminate option C. Now we need to figure out this bearing to narrow down um between the rest of the answer traces. If we think about this bearing and the bearing, the B that we drew in red is gonna be made up of two things. We have this 257 degrees that we already drawn plus this little angle inside of our triangle, we can call that the OK. So our angle theta B, the angle of the bearing that we're looking for is gonna be equal to 257 degrees plus this angle theta. Hm. Now this angle theta we can use the law of signs to figure it out because we know the opposite side and then we know the side and the opposite angle for another pair in our triangle. So let's go ahead and use the law of signs to find data. I recall that the law of science tells us that the ratio of the sine of an angle divided by its corresponding side is equal for all of the sides and angles in a triangle. So sine A divided by A is equal to sine B divided by B. In this case, we'll let a be that angle theta that we're looking for. And so the opposite side will be 50 MPH. So this is gonna give us sign of the divided by 50 MPH is gonna be equal to sine of B. So we wanna use that other angle. We know 132 degrees divided by the opposite side, which is that ground speed we just calculated. So we're dividing by 584.6385 MPH, we can multiply both sides by 50 MPH. We're gonna have that sign of that angle. Theta that we're looking for is gonna be equal to 50 MPH, multiplied by sign of 132 degrees divided by 584.6385 MPH. The unit of MPH will divide out. We can work out the right hand side, then take the inverse sign. And when we do that, we get that this angle, theta that we were looking for is gonna be approximately equal to 3.644 degrees. OK? So we have this angle theta, but we have to remember we're looking for the bearing and we've already said from our diagram that, that bearing is gonna be 257 degrees plus this angle theta that we found from our triangle. Now, we have that value. So we know that the angle of the bearing is gonna be 257 degrees plus 3.6 or four degrees, which gives us a final value of the bearing of 260.644 degrees. Approximately, right. So we have the ground speed, we now have the resulting bearing. We can get back to answering this question. If we take a look at our answer choices, we can see that the correct answer is going to be. Option B the resulting bearing is 261 degrees with the ground speed of 585 MPH. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Bearing and Direction in Navigation

Vector Addition of Velocities

Trigonometric Resolution of Vectors

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