Answer:
The wind pushed the plane [tex]65.01[/tex] miles in the direction of [tex]45.56 ^{\circ}[/tex] East of North with respect to the destination point.
Step-by-step explanation:
Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the starting point in the direction of 35° E of South as shown in the figure.
So, we have |OD|=250 miles and |OD'|=230 miles.
Vector [tex]\overrightarrow{DD'}[/tex] is the displacement vector of the plane pushed by the wind.
From figure, the magnitude of the required displacement vector is
[tex]|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)[/tex]
and the direction is [tex]\alpha[/tex] east of north as shown in the figure,
[tex]\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)[/tex]
From the figure,
[tex]|AB|=|OA-OB|[/tex]
[tex]\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|[/tex]
[tex]\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|[/tex]
[tex]\Rightarrow |AB|=45.52[/tex] miles
Again, [tex]|PQ|=|OP-OQ|[/tex]
[tex]\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|[/tex]
[tex]\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|[/tex]
[tex]\Rightarrow |PQ|=46.42[/tex] miles
Now, from equations (i) and (ii), we have
[tex]|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01[/tex] miles, and
[tex]\tan \alpha=\frac{|46.42|}{|45.52|}[/tex]
[tex]\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}[/tex]
Hence, the wind pushed the plane [tex]65.01[/tex] miles in the direction of [tex]45.56 ^{\circ}[/tex] E astof North with respect to the destination point.
