Answer
a) The distance she has to fly from Manhattan to get back to Lincoln is $189km$
b) She must fly in the direction $10.5^{\circ}$ West of North
Work Step by Step
Our coordinate system is such that (+)ve y-direction is along the North and (+)ve x-direction is along the East.
Let $A,B,C$ be the displacement vectors and their sum is the resultant vector $R$
$A=147km, 85^{\circ} from North$
$B=106km, 167^{\circ} from North$
$D=166km, 235^{\circ} from North$
Now,$R=A+B+C$.
So,$R_{x}=A_{x}+B_{x}+C_{x}$ and $R_{y}=A_{y}+B_{y}+C_{y}$
As the angles are measured from the North,we have
$R_{x}=147sin(85^{\circ})+106sin(167^{\circ})+166sin(235^{\circ})=34.3km$
and,$R_{y}=147cos(85^{\circ})+106cos(167^{\circ})+166cos(235^{\circ})=-185.7km$
Since ,we are interested in the return trip from Manhattan to Lincoln,we are interested in the negative of $R$ i.e, $R'=-R=(-34.3km)i+(185.7km)j$ and $R'$ lies in the 2nd quadrant
The magnitude of $R'$ is
$|R'|=\sqrt ((-34.3)^2+(185.7)^2)=189km$
and its direction is given by
$tan(\theta)=\frac{R'_{x}}{R'_{y}}=\frac{-34.3}{185.7}$ [since,angles are meaured from the North]
$\theta=tan^{-1}(\frac{-34.3}{185.7})=-10.5^{\circ}$
a) The distance she has to fly from Manhattan to get back to Lincoln is $189km$
b) She must fly in the direction $10.5^{\circ}$ West of North
