The direction of the displacement is 16.2⁰ north east.
The given parameters;
initial displacement, x₁ = 53 m north
final displacement, x₂ = 45 m east
The magnitude of the resultant displacement is calculated as follows;
form the diagram uploaded,
[tex]R^2 = 53^2 + 45^2 - 2(53 \times 45) cos(150)\\\\R^2 = 4834 + 4130.82\\\\R = \sqrt{8964.82} \\\\R = 94.68 \ m[/tex]
The direction of the displacement is calculated as;
[tex]\frac{53}{sin(\theta)} = \frac{94.68}{sin(150)} \\\\\frac{53}{sin(\theta)} = 189.36\\\\sin(\theta) = \frac{53}{189.36} \\\\sin(\theta) = 0.2798\\\\\theta = sin^{-1} (0.2798)\\\\\theta = 16.2^0 \ north \ east[/tex]
Thus, the direction of the displacement is 16.2⁰ north east.
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