A person moves $20$m north and then A person moves $20$m north and then he moves $30$m south west and then finally he moves $10$m east. Find the magnitude and direction of resultant displacement. 
My Attempt :

I have drawn the figure, but not sure whether it is correct. Moreover, I couldn't get further! 
 A: You're on the right track but not quite there.
The process of drawing the net displacement by putting the tail of the next vector at the head of the previous one is correct.
But the length of the second vector I know is incorrect because its head will not land on the $x$ axis. (It actually lands a little bit past the $x$ axis.) So you made an assumption that it would, and it's not correct.
Here's how I'd tackle it.
Your first vector is $\vec{v_1} = 20\hat{y}$, measuring in meters. (The variable $\hat{y}$ -- I call it "y-hat" -- is a unit vector in the $y$ direction.)
The second vector has length $30$ (meters) pointing southwest. So I'd separate this into components using $\vec{v_2} = 30(\cos(225^{\circ})\hat{x} + \sin(225^{\circ})\hat{y}) = -(30/\sqrt{2})\hat{x} - (30/\sqrt{2})\hat{y}.$
The last vector is easier; I'll let you try that one.
Then, add all of the $\hat{x}$ components and all of the $\hat{y}$ components separately.
Once you have that (which will be something like $\vec{V} = a\hat{x} + b\hat{y}$) then you can get the magnitude (length) as $R = \sqrt{a^2 + b^2}$ and the angle as $\theta = \tan^{-1}(b/a)$, where you'll need to look at the signs of $a$ and $b$ to determine the quadrant.
Can you take it from here?
A: The assumption that $B$ and $O$ share the same $y$ coordinate seems to be not correct.
Guide:
position $A$ is $(0, 20)$.
position $B$ is $(0,20) - 30 (\sin 45^\circ, \cos 45^ \circ)$.
