How do I do this trigonometry vector calculation?

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A pilot needs to calculate the correct heading to fly due north while accounting for a 30 km/h wind from the southeast, with the plane's speed in still air being 100 km/h. The initial calculations yielded an angle of 9.93 degrees and a speed of 123 km/h, but the reference answers indicate 12.24 degrees and 119 km/h. The discussion emphasizes the importance of correctly visualizing the vector triangle, ensuring that the plane's velocity vector is oriented towards true north. It highlights the necessity of drawing vectors head-to-tail and accurately determining angles based on the wind's direction. Acknowledging the correct approach can significantly clarify the solution process for similar physics problems.
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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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