Answer
$38.26\ m$
Work Step by Step
Given
Magnitude of each vector = 50 m
As the vectors $\vec{a}, \vec{b}\ and\ \vec{c}$ of magnitude 50 m are making angles of $30^{\circ} , 195^{\circ}\ and\ 315^{\circ}$ with the positive x axis, then the vectors in unit vector notation is
$\vec{a} = 50m\ \cos30^{\circ} \hat{i}+50m\ \sin30^{\circ} \hat{j}$
$\vec{a} = 43.30\ m\ \hat{i}+ 25\ m\ \hat{j}$
$\vec{b} = 50m\ \cos195^{\circ} \hat{i}+50m\ \sin195^{\circ} \hat{j}$
$\vec{b} = -48.30\ m\ \hat{i}+(-12.94\ m)\ \hat{j}$
$\vec{c} = 50m\ \cos315^{\circ} \hat{i}+50m\ \sin315^{\circ} \hat{j}$
$\vec{c} = 35.35\ m\ \hat{i}+(-35.35\ m)\ \hat{j}$
The sum of three vector $\vec{a}, \vec{b}\ and\ \vec{c}$ is given by
$\vec{a}+\vec{b}+\vec{c} = 43.30\ m\ \hat{i}+ 25\ m\ \hat{j}-48.30\ m\ \hat{i}-12.94\ m\ \hat{j}+35.35\ m\ \hat{i}-35.35\ m\ \hat{j}$
$\vec{a}+\vec{b}+\vec{c}=(30.35\hat{i}-23.3\hat{j})\ m $
The magnitude of the vector is given by
$\mid{\vec{a}+\vec{b}+\vec{c}}\mid = \sqrt(30.35^2+(-23.3)^2)$
$\mid{\vec{a}+\vec{b}+\vec{c}}\mid=38.26\ m$
