How would you solve this question:
A plane that can fly at 250 km/h wishes to reach an airport that has a bearing of 25° W of N from its present location. If there is a 50.0 km/h wind blowing directly to the west what should be the heading of the plane?
What will be its ground speed?
Thanks
Sounds like someone wasn’t paying attention in class.
Thanks for nothing. For your information, I am not in school, and this is not homework. If you don’t want to help, that’s fine. But don’t waste my time.
No offense, but you’re going to at least have to come up with a plausible explanation for what this is, before we’ll believe that.
It’s an example of a homework problem I am trying to walk someone I am tutoring through. The problem comes from this site. I have the answer to the question. I have done the work, but I think I am making a mistake somewhere that I cannot figure out. Hence, the question. Honestly, if I were trying to cheat on homework, do you really think I would copy the question verbatim, then post it to a public message board indexed by google? Really?
In the future, if you don’t believe what I say, don’t bother responding. I’d think the question justifies the interrogation.
Do we get a cut of your fee?
You may not be aware of this, but this board has had instances of youngsters seeking homework assistance who post in a very similar manner as what you just did and anyone’s username is woefully ineffectual at providing an insight into someone’s inherent integrity.
I apologize wasting your time.
Sure, half of free is free. Where should I mail your check? Thanks again for all your help guys. I really appreciate it.
I’m doing this from memory, and based on your description, what you have is a triangle with the bottom length 50 and the left line 250. Crap this is hard to describe
cosine law: a^2 = b^2 + c^2 -2 bc*cos(beta)
beta in this case is 90-25 = 65
You want a line that goes off to the left at an angle of 25 degrees, so draw that. From that start point, draw a line to the left that’s 50 long. Then from the end of that short line, draw a line up to the left that is 250 long. Then end of the 250 should touch the end of your original “desired” vector. Thus vector addition.
From File:Triangle with notations 2.svg - Wikipedia
b is 250
c is 50
a is your ground speed
you are looking to find alpha
I get a ground speed of 266.99, but I’m currently finding the wrong heading.
There’s a pretty good resource here.
So my final answer is that the plane needs to head 14.5 degrees west of north, and its ground speed will be 266.9896.
Anyone else?
As I said, solving the vector diagram requires using the cosine law
a^2=x^2+c^2-2xc*cos(alpha)
where alpha = 90 - 25 (your known angle)
requires using the quadratic equation to solve for x, where x is your ground speed
then use the cosine law again using the known value of x
cos(theta) = [a^2 + b^2 - x^2] / 2ab
the angle the plane flies in then theta - 90
don’t forget to switch between radians and degrees.
Thanks to all who responded with answers. I realized my mistake, so I figured out the answer. Also, I think it’s easier to solve with the law of sines, but both way work. Thanks again.
I’m not sure you have enough info to use Sine Law.
Was my answer correct?
ETA Never mind, the internet told me I was right.
Your answers were correct. With the law of sines, you just use the speed and wind to find the other angle, then subtract the sum from 180. Upon reflection, your way is easier. Thanks again.
I just drew a diagram and got 50cos(25) = 250sinX, where X is the angle from the desired ground path (-25deg, not North).
For ground speed, it’s 250cos(x) + 50sin(25).
Why are you guys using a law of sines/cosines at all?
Don’t go off on a tangent
Because around here we obey the law!
In other words, I have no idea what you did or why 50cos(25) = 250sin(X)
On further review, the Sin Law is way easier, since the triangle has three sides: 250, 50, and x
It also has three angles: 65, theta, and 180-65-theta
OK, first of all, North does absolutely nothing for us, so don’t even draw that line.
Draw a three-line diagram. Pretend it’s a clock. The airport is at 12, the plane is flying toward 1 and the wind is coming in from 4. The plane is the minute hand, since it’s going faster. The magnitude is 250, right? The wind is the hour hand and it’s magnitude is 50.
In order for the plane to get to 12, the horizontal component of the wind has to equal the horizontal component of the plane.
The wind’s component is 25deg below the 3 o’clock point. So the horizontal component is 50cos(25). The plane’s component is 250sin (x) if X is the angle between the airport (12 o’clock) and the plane’s heading (1 o’clock).
Like I said, they’re equal. So 50cos(25)=250sin(X). Once you’ve got X, you know can add/subtract it from the “25 W of N” heading to find the plane’s heading. Finally, you just add the two vertical components together, which you can get by flipping the sin and cos functions, to get the groundspeed.
See? No law of Sines or anything. Just two triangles and the realization that they share a common side. It helps to realize that North is arbitrary and only useful as a label, not as a mathematical value.
