Question

A particle undergoes two displacements, measured from the positive x-axis, with counterclockwise positive. The first has a magnitude
of 11 m and makes an angle of 69
◦
with the
positive x axis. The resultant displacement has a magnitude of 10 m directed at an angle
of 120
◦
from the positive x axis. Find the magnitude of the second displacement.

Answer 1

I know I need to find B=\[\sqrt{bx^2+by^2}\] and Bx=Cx-Ax and By=Cy-Ay

Answer 2

help plz??

Answer 3

break both the resultant and the first displacement into components of x and y. Subtract them into the components of displacement B. Use the Pythagorean theorem and you should get:
<b>=9.09m
use tangent fot the angle to get:
angle=maybe 7.4 degrees

Answer 4

magnitude of the second displacement (b)=
b^2=11*11+10*10-10*11*cos(51)
b^2=221-81.6=139.36
b=11.8