Aspelunker is surveying a cave. she follows a passage that takes her a distance 184 m straight west, then a distance 220 m in a direction 30.0 degrees south of east, then a distance 104 m at 80.0 degrees east of north. after a fourth unmeasured displacement she find herself back where she started. use the method of components to find the magnitude and direction of the fourth displacement.

Refer to the diagram shown below.

Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.

The first distance is 184 m west. It is represented by
d₁ = -184 i

The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j

The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) =  102.42 i + 18.06 j

Let the fourth distance be 
d₄ = a i + b j

Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.

The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95

The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94

The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m

The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°

The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.