An aeroplan heads in a direction of S30ºW at a speed of 300km/h. The wind b**w at a speed of 50km/h in the direction of S20ºE . What is the

I do this one a little differently from Melody.......but the answers are the same.

First...note that S30°W = 240°    .....and S20°E = 290°

So we have two vectors here with the following components....

<300cos240°, 300sin240° >   and < 50cos290°, 50sin290° >

And adding the first to components of the two vectors we get.....300cos240° + 50cos240° = -132.8989928337

And adding the second components of the two vectors we get.....300cos290° + 50cos290° = -306.7922521745

And to find the magnitude of the resultant vector, we have.......

√[( -132.8989928337)²  + ( -306.7922521745)² ] = √(111783.6282905138439343) ≈ 334.34 km/h

And to find the resultant heading, we have

tan-1 [(-306.7922521745) / ( -132.8989928337)] = tan-1 (2.308461829792775) = 66.578275555105°

And to find the resulting heading, we have.... S(90 - 66.578275555105)°W  ≈ S(23.42)° W

I actually like this method better because I always have a little trouble visualizing the "parallelogram law."  In effect, I just let the math "do the work" for me and I don't have to worry about drawing any pictures.

But, it's purely a matter of preference....!!!!

Mmm interesting.

I have drawn this to scale.

Use Cosine Rule to find R

To get the 130degrees I used the fact that interior angles on parallel lines are Supplementary.

$$\\R^2=50^2+300^2-(2*50*300*cos130^0)\\
R^2=111783\\
R=334.34 \;km/hour$$

Now find the angle at the top of the triangle - Use Cosine Rule again.

I forgot to label it on the diagram but i have called it $$\alpha$$  in the working.

$$\\50^2=300^2+334.34^2-2*300*334.34*Cos(\alpha)\\\\
\alpha=acos\left(\frac{334.34^2+300^2-50^2}{2*300*334.34}\right)\\\\
\alph=6.578^0\\\\
therefore\\\\
\theta=30-6.578=23.42^0$$

$$So the plane is travelling S 23.4$^0$ W at 334km/hour$$

Note: I have not checked my working.    Any questions?  Just ask.