Actual displacement that he required to move
[tex]d_1 = 5.4 km [/tex] towards North
Displacement that he moved due to snow is
[tex]d_2 = 8.1 km[/tex] at 47 degree North of East
now in vector component form we can say
[tex]d_1 = 5.4 \hat j[/tex]
[tex]d_2 = 8.1 cos47 \hat i + 8.1 sin47 \hat j[/tex]
[tex]d_2 = 5.52 \hat i + 5.92 \hat j[/tex]
now the displacement that is more required to reach the destination is given as
[tex] d = d_1 - d_2[/tex]
[tex]d = 5.4\hat j - (5.52 \hat i + 5.92\hat j)[/tex]
[tex]d = -5.52 \hat i - 0.52 \hat j[/tex]
so the magnitude of the displacement is given as
[tex]d = \sqrt{5.52^2 + 0.52^2}[/tex]
[tex]d = 5.54 km[/tex]
its direction is given as
[tex]\theta = tan^{-1}\frac{0.52}{5.52} = 5.38 degree[/tex]
so it is 5.54 km towards 5.38 degree North of West.