While surveying a cave, a spelunker follows a passage 190m straight west, then 210m in a direction 45.0east of south, and then 270m at 30.0 east of north. after a fourth unmeasured displacement, she finds herself back where she started. i need to use vector components to find the magnitude and direction of the fourth displacement.

The first thing you need to consider all the displacements in y direction.
S1 has no displacement in y direction.
S2 has a displacement of (-210 sin 45)
S3 has a displacement of (270 sin30)
but
S1 + S2+ S3+ S4 = 0 0 + (-210 sin 45) + (270 sin30) + Sy=0 Sy= +105 *sqrt2 - 135 = 148.05 -135 = 13.05
 
You need to consider also the displacement in x direction
S1 has a displacement of -190 m
S2 has a displacement of 210 cos45
S3 has a displacement of 270 cos 30 S1 + S2 + S3 + S4 = 0 -190 + 105 * sqrt2 + 135* sqrt3 +S4 = 0 S4 =190 - 148.05 - 233.55= - 191.6
Final displacement would be the result of 191.6i - 13.05j
I hope this can help

Less force and energy was applied

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