Answer
a) The vector-addition diagram roughly to scale is shown in figure.
b) For the explorer to return back to the shelter ,he/she must take $49steps$, $76^{\circ}$ South of East (or,$14^{\circ}$ East of South).
Work Step by Step
Let East be along the (+)ve x-direction and North be along the (+)ve y-direction
Let $A,B,C$ be the displacement vectors and $R$ be the resultant vector.
$A=40 steps$ ,NorthEast
$B=80steps$ ,$60^{\circ}$ North of West
$C=50steps$ ,due South
Now,$R=A+B+C$
So,
$R_{x}=A_{x}+B_{x}+C_{x}$ and $R_{y}=A_{y}+B_{y}+C_{y}$
$R_{x}=40cos(45^{\circ})+80cos(120^{\circ})+50cos(270^{\circ})=-11.7$
and,$R_{y}==40sin(45^{\circ})+80sin(120^{\circ})+50sin(270^{\circ})=47.6$
We are interested in returning the explorer back to the shelter, so we are interested in the negative of $R$ i.e, $R'=-R=11.7i+(-47.6)j$ and $R'$ lies in the 4th quadrant
The magnitude of $R'$ is
$|R'|=\sqrt ((11.7)^2+(-47.6)^2)\approx49$
and its direction is
$tan(\theta)=\frac{R'_{y}}{R'_{x}}=\frac{-47.6}{11.7}$
$\theta=tan^{-1}(\frac{-47.6}{11.7})=-76.2^{\circ}\approx-76^{\circ}$
a) The vector-addition diagram roughly to scale is shown in figure.
b) For the explorer to return back to the shelter ,he/she must take $49steps$, $76^{\circ}$ South of East (or,$14^{\circ}$ East of South).
