A man has to travel 50 m due north, 40 m due east, and 20 m due...

Let the starting point be the origin $(0,0)$. Taking East along the positive x-axis and North along the positive y-axis, the individual displacements are:

$\vec{d}_1 = 50 \hat{j}$

$\vec{d}_2 = 40 \hat{i}$

$\vec{d}_3 = -20 \hat{j}$

The net displacement vector is:

$\vec{D} = \vec{d}_1 + \vec{d}_2 + \vec{d}_3 = 40 \hat{i} + (50 - 20) \hat{j} = 40 \hat{i} + 30 \hat{j}$

The magnitude of the displacement is:

$|\vec{D}| = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = 50 \text{ m}$

The angle $\theta$ made with the East direction is:

$\tan \theta = \dfrac{y}{x} = \dfrac{30}{40} = \dfrac{3}{4}$

$\theta = \tan^{-1}(3/4)$ North of East.