Find the magnitude of the third displacement?

Question

A particle undergoes three displacements. The first has a magnitude of 15 meters and makes an angle of 29 degrees with the positive x axis. The second has a magnitude of 8.9 meters and makes an angle of 131 degrees with the positive axis. After the third displacement the particle returns to its initial position. Find the magnitude of the third displacement in meters.

Answer 1

Let the particle be at origin initially, the coordinates of the particle after 1'st displacement be (x1,y1) & after second displacement be (x2,y2)

Now, x1 = 15cos29 & y1 = 15sin29

and, x2 = x1 + 8.9cos131 & y2 = y1 + 8.9sin131

Thus, position of particle after 2nd displacement,

         (x2,y2) = (7.28,13.99)

The third displacement will bring the particle to its initial position (0,0)

Therefore, the maginitude of third displacement is given by

sqrt [(0-7.28)² + (0-13.99)²]

= 15.77 meters