Correct option is (c) tan-1 (1/5)
Because the pilot maintains an eastward heading, the engines still push the plane due east. So, vplane = 100m/seastward. But the wind blows at vwind = 20m/s northward, thereby knocking the plane off course by angle ϕ.
In the figure, we know the two legs but not the hypotenuse. Specifically, in units of m/s, the opposite leg has length ho = 20, while the adjacent leg has length ha = 100. Therefore
tanϕ = \(\frac{h_o}{h_a} = \frac{20}{100} \)
Thus ϕ = tan−1\( \frac15\)