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
![\vec{W}=-32\hat{j}](https://img.qammunity.org/2023/formulas/mathematics/college/edto08axothj43t7ep8dkwlunjgafyeub4.png)
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
![\vec{R}=140\cos 24\hat{i}+140\sin 24\hat{j}](https://img.qammunity.org/2023/formulas/mathematics/college/g4li9lmpmnykbijzzx5y6emos6dmffc43f.png)
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
![\vec{T}=T\cos \theta\hat{i}+T\sin \hat{j}](https://img.qammunity.org/2023/formulas/mathematics/college/hgf94kdfu5yjsbctegttrusex2xpo888z6.png)
Now, we know that the resultant is
![\vec{R}=\vec{T}+\vec{W}](https://img.qammunity.org/2023/formulas/mathematics/college/x2i51egzj1ju0sm81xgtxgmwqticsyxjzy.png)
then
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/rrsfrrh1aq5jpv08m1n8c9c64h53vhwk0g.png)
Since the unit vector i and j are independent this gives us two equations
![\begin{gathered} 140\cos 25=T\cos \theta \\ 140\sin 25=T\sin \theta-32 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rwyeuqjamrew9vpg816ex0mpwy8k77xozm.png)
From the first equation we have that
![T=(140\cos 25)/(\cos \theta)](https://img.qammunity.org/2023/formulas/mathematics/college/i7dlquz36jqulnsnf40xe59zz9fs8kaw7o.png)
Plugging the value of T in the second equation we have
![140\sin 25=((140\cos25)/(\cos\theta))\sin \theta-32](https://img.qammunity.org/2023/formulas/mathematics/college/d9eup29ypqssidraq6nzmf6rf73rffytvg.png)
Now we need to solve this equation for theta.
![\begin{gathered} 140\sin 25=((140\cos25)/(\cos\theta))\sin \theta-32 \\ 140\sin 25+32=(140\cos 25)(\sin \theta)/(\cos \theta) \\ (\sin\theta)/(\cos\theta)=(140\sin 25+32)/(140\cos 25) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3sq8n39u9kaw4cf4okwkaa4pzl4e5vgcca.png)
Now we have to remember that
![(\sin\theta)/(\cos\theta)=\tan \theta](https://img.qammunity.org/2023/formulas/mathematics/college/e17mk8ea1djx16tadkvpgixtuf0eg5ozdx.png)
hence
![\begin{gathered} \tan \theta=(145\sin 25+32)/(145\cos 25) \\ \theta=\tan ^(-1)((145\sin 25+32)/(145\cos 25)) \\ \theta=35.7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d03kbsffk2wvbkon2tqfgw5n5bq7qnscje.png)
Therefore Diane has to direct the airplane at an angle of 35.7° North of east.