How do I do this trigonometry vector calculation?

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Homework Help Overview

The problem involves a pilot attempting to fly due north while accounting for a wind blowing from the southeast. The pilot's plane has a specified airspeed, and the task is to determine the necessary heading and ground speed relative to the wind conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the vector triangle representing the plane's motion and the wind's influence. There are questions about the correctness of the original poster's triangle scheme and the angles involved. Some participants suggest re-evaluating the angles and the direction of the vectors.

Discussion Status

The discussion is ongoing, with participants providing insights into the vector relationships and the importance of correctly visualizing the problem. There is acknowledgment of different interpretations of the angles involved, and some guidance has been offered regarding the vector addition principle.

Contextual Notes

There are indications of confusion regarding the angles and the setup of the triangle, as well as the original poster's difficulty in visualizing the problem correctly. The original poster expresses a realization of their misunderstanding after receiving feedback.

Jan Berkhout
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Homework Statement


A pilot wishes to fly at maximum speed due north. The plane can fly at 100km/h in still air. A 30km/h wind blows from the south-east.
Calculate:
a) The direction the plane must head to fly north.
b) Its speed relative to the ground.

Homework Equations


Sine Rule: a/SinA=b/SinB=c/SinC
Cosine Rule: a2=b2+c2-2acCosA

The Attempt at a Solution


I attached a photo because I didn't know how to do maths on here.
The answer I got was 9.93 degrees and 123km/h

vectors.png

https://pasteboard.co/HwvgTjt.png
The answers from the sheet I'm working off say 12.24 degrees and 119km/h. And I don't understand how their way is correct. To get the angle they did: sin45/100=sinθ/30 If 100 is the planes speed it wouldn't be opposite to the angle of the direction of the wind.
 

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Your triangle scheme is not correct. According to your scheme, the plane final speed (which is represented by the side b=(AC)) isn't towards north but somewhere between North-Northeast. Make again the triangle in such a way that the side b is vertical towards north and has magnitude unknown which we wish to find, side a is the direction of wind (from south east to northwest, makes 45 degrees angle with side b) and side c has magnitude 100 and direction unknow theta which we wish to find.
 
It is useful to memorise a vector equation that your relative motion triangles must always obey:

##v_b\ =\ v_a\ +\ v_{b\ rel\ a}##

and remember that to show this vector addition, the two vectors being summed need to be drawn head-to-tail.

You are making things difficult for yourself if you don't draw North as heading vertically up the page.
 
Jan Berkhout said:
I attached a photo because I didn't know how to do maths on here.
The answer I got was 9.93 degrees and 123km/h

In your diagram, you can't assume angle B is 135 degrees. Angle B is the angle between the pilot's unknown direction and the direction of the wind.

The given information let's us deduce the angle between due North and the direction of the wind. So we can deduce angle C.
 
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Thank you all! I approached this question wrong and I just couldn't see a different way to look at until you showed me! This often happens to me with physics questions, Cheers!
 
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