Friday

April 18, 2014

April 18, 2014

Posted by **Pauline** on Friday, March 30, 2007 at 9:41am.

I drew the following diagram.

OX is the x-axis, and OY is the y-axis

OB is the desired vector you want to go

so angle(YOB)=40º, it has a magnitude of 500.

Let OA be the vector the plane has to fly, AB is the vector representing the direction and velocity of the wind.

(point A is to the right of B and slightly downwards so it makes a 10º angle with the horizonatal)

Using a bit of basic geometry it should be easy to follow that angle(OBA)=120º

If we let the time taken be t hours, we can label the maginitude of AB as 56t and the magnitude of OA as 350t

Using the cosine law I got the equation

(350t)^2 = 500^2 + (56t)^2 - 2(500)(56t)cos 120º

or

119364t^2 - 28000t - 250000=0

check my arithmetic but by the quadratic formula I got t=1.56 or (a negative t, which we reject)

So the time taken is 1.56 hours

=1.56*60 minutes

=94 minutes to the nearest minute.

Using the Sine Law, Sin(BOA)/(56*1.56) = sin 120/(350*1.56)

sin(BOA)=.1385

angle(BOA)=7.96º

so the direction should be 40º + 7.96º

or appr 48º

Setting up a similar geometrical figure you should be able to do the second part yourself.

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