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Degger [83]
3 years ago
5

LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the

Mathematics
1 answer:
Semmy [17]3 years ago
5 0

Answer:

The wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} 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 \overrightarrow{DD'} is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)

and the direction is \alpha east of north as shown in the figure,

\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)

From the figure,

|AB|=|OA-OB|

\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|

\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|

\Rightarrow |AB|=45.52 miles

Again, |PQ|=|OP-OQ|

\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|

\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|

\Rightarrow |PQ|=46.42 miles

Now, from equations (i) and (ii), we have

|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01 miles, and

\tan \alpha=\frac{|46.42|}{|45.52|}

\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}

Hence, the wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} E astof North  with respect to the destination point.

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88.88% probability that it endures for less than a year and a half

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

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