EXPLANATION
Let's represent the situation on a graph, we know the plane moves at 150 miles per hour at 100 degrees east of north; but, if we measure from the positive x-axis, this corresponds to a -10 degree angle.
The velocity's x-component is thus as follows:
[tex]150\cdot\cos (-10)=147.72\text{mph}[/tex]
The velocity's y-component is thus as follows:
[tex]150\cdot\sin (-10)=-26mph[/tex]
Therefore, the velocity vector is as follows:
V = < 147.7 mph, -26 mph >
Now, we need to represent the wind vector, applying the same reasoning than above:
x-component: 50*cos(225°) = -35.4 mi/hr
y-component: 50mi/hr*sin(225°) = -35.4 mi/hr
Therefore the wind vector is as follows:
W = < -35.4 mi/hr, -35.4 mi/hr>
Then, adding both vectors:
resultant V + W = (147.7 -35.4 , -26 -35.4)
resultant V + W = (112.3mph, -61.4mph)
Finally, we need to get the direction of the vector:
θ = tan^-1 (y/x)
Plugging in the numbers into the expression:
[tex]\theta=\tan ^{-1}(\frac{-61.4}{112.3})[/tex]
Computing the argument:
[tex]\theta=\tan ^{-1}(\frac{-61.4}{112.3})=-28.7\text{ degre}es[/tex]
In conclusion, the direction of the plane is 28.7 degrees south of east
