AP Physics
posted by Annamaria .
While exploring a cave, a spelunker starts at the entrance and moves the following distances: 75.0 m north, 250 m east, 210 m at an angle 30.0° north of east, and 150 m south. Find the resultant displacement from the cave entrance.
I know I have to use the formulas Ax = A cos è, Ay = A sin è, A = sq rt((Ax)^2 + (Ay)^2), and è = tan^(1) (Ay/Ax). But I'm not sure whether I should calculate 125*sin(30) on the calculator because I get a negative number.

è is theta

125sin(30) is 125x0.5 = 62.5
For this sort of problem you need to draw it out on paper and so hard to show on here.
210 m sin(30) + 75 m= 105 m gives the total displacement north. After this he moves south by 150 m so his displacement north is
105m+150m +75 m = 30 m
[North is the +ve direction)
His displacement east is
250 m + 210 cos 30 = 250 m +182 m = 432 m
so the displacement angle is
tan^1(30/432) = 3.97 deg (North of east)
and the displacement is
Sqrt(30^2 + 432^2) = 433 m
But check my maths!! 
Thanks