Answer
a) To return to the oak tree, one has to walk $818m$ , $15.8^{\circ}$ West of South.
b) From the vector-addition diagram, our calculation in part (a) is resonable.
Work Step by Step
Let us choose our coordinate sytem in such a way that North is along the (+)ve y-direction and East is along the (+)ve x-direction.
Let $A,B,C$ and $D$ be the displacement vectors where $D$ is the unknown.
$A=0.825km$, due South
$B=1.25km$, $30.0^{\circ}$West of North
$C=1.00km$, $32.0^{\circ}$North of East
Since, we head back to the old oak tree ,the sum of the displacement vectors is 0 i.e,
$A+B+C+D=0$
$=>D=-(A+B+C)$
Thus, $D_{x}=-(A_{x}+B_{x}+C_{x})$ and $D_{y}=-(A_{y}+B_{y}+C_{y})$
$D_{x}=-(0.825cos(-90^{\circ})+1.25cos(90^{\circ}+30.0^{\circ})+1.00cos(32.0^{\circ})=-0.223km$
$D_{y}=-(0.825sin(-90^{\circ})+1.25sin(90^{\circ}+30.0^{\circ})+1.00sin(32.0^{\circ})=-0.787km$
So, $D=(-0.223km)i+(-0.787km)j$ and it lies in the 3rd quadrant
The magnitude of $D$ is
$|D|=\sqrt ((-0.223)^{2}+(-0.787)^{2})=0.818km=818m$
and the direction of $D$ is given by
$tan\theta=\frac{D_{y}}{D_{x}}=\frac{-0.787}{-0.223}$
$\theta=tan^{-1}(\frac{0.787}{0.223})=74.2^{\circ}$
Since ,$D$ is in the 3rd quadrant, it is off by $180^{\circ}$.
Therefore, the direction of $D$ is $270^{\circ}-(180^{\circ}+74.2^{\circ})=15.8^{\circ}$West of South
a) To return to the oak tree, one has to walk $818m$ , $15.8^{\circ}$ West of South.
b) The vector-addition diagram is sketched roughly to scale .We see that the final displacement vector $D$ from this sketched agrees with the value of $D$ from (a)
