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Sagot :
We know that the wind is blowing from north to south with a speed of 32 km per hour, let's call the wind vector W, then it can be written as
[tex]\vec{W}=-32\hat{j}[/tex]We also know that Diane wants the plane to fly north of east at an angle of 25° and a speed of 140 km/h. This means that the resultant should be
[tex]\vec{R}=140\cos 24\hat{i}+140\sin 24\hat{j}[/tex]Now, we would like to know at what angle Diane should fly tha plane. To find this we will introduce a vector T whose magnitud and direction are unknown.
Then
[tex]\vec{T}=T\cos \theta\hat{i}+T\sin \hat{j}[/tex]Now, we know that the resultant is
[tex]\vec{R}=\vec{T}+\vec{W}[/tex]then
[tex]\begin{gathered} 140\cos 25\hat{i}+140\sin 25\hat{j}=T\cos \theta\hat{i}+T\sin \theta\hat{j}+(-32\hat{j}) \\ 140\cos 25\hat{i}+140\sin 25\hat{j}=T\cos \theta\hat{i}+(T\sin \theta-32)\hat{j} \end{gathered}[/tex]Since the unit vector i and j are independent this gives us two equations
[tex]\begin{gathered} 140\cos 25=T\cos \theta \\ 140\sin 25=T\sin \theta-32 \end{gathered}[/tex]From the first equation we have that
[tex]T=\frac{140\cos 25}{\cos \theta}[/tex]Plugging the value of T in the second equation we have
[tex]140\sin 25=(\frac{140\cos25}{\cos\theta})\sin \theta-32[/tex]Now we need to solve this equation for theta.
[tex]\begin{gathered} 140\sin 25=(\frac{140\cos25}{\cos\theta})\sin \theta-32 \\ 140\sin 25+32=(140\cos 25)\frac{\sin \theta}{\cos \theta} \\ \frac{\sin\theta}{\cos\theta}=\frac{140\sin 25+32}{140\cos 25} \end{gathered}[/tex]Now we have to remember that
[tex]\frac{\sin\theta}{\cos\theta}=\tan \theta[/tex]hence
[tex]\begin{gathered} \tan \theta=\frac{145\sin 25+32}{145\cos 25} \\ \theta=\tan ^{-1}(\frac{145\sin 25+32}{145\cos 25}) \\ \theta=35.7 \end{gathered}[/tex]Therefore Diane has to direct the airplane at an angle of 35.7° North of east.
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