here we need to write the two components of the displacement and then we need to add them
first displacement is given as
[tex]d_1 = 75 km[/tex] 30 degree West of North
[tex]d_1 = - 75 sin30 \hat i + 75 cos 30 \hat j[/tex]
[tex]d_1 = - 37.5 \hat i + 64.9\hat j[/tex]
Other displacement is given as
[tex]d_2 = 155 km[/tex] 60 degree due East of North
[tex]d_2 = 155 sin60 \hat i + 155 cos60 \hat j[/tex]
[tex]d_2 = 134.2 \hat i + 77.5 \hat j[/tex]
now we need to find the net displacement
[tex]d = d_1 + d_2[/tex]
[tex]d = (-37.5 + 134.2)\hat i + (64.9 + 77.5)\hat j[/tex]
[tex]d = 96.7\hat i + 142.4\hat j[/tex]
so it is given as
[tex]d = \sqrt{96.7^2 + 142.4^2}[/tex]
[tex]d = 172.1 km[/tex]
and its direction is given as
[tex]\theta = tan^{-1}\frac{96.7}{142.4} = 34.1 degree[/tex]
so it will displace by 172.1 km at 34.1 Degree East of North
