May you please help me solve this physics question?
Determine the angle that the vector sum of the displacements Δd1 = 6.2 m [58.0° S of W], Δd2 = 6.8 m [S] and Δd3 = 28.0 m [17.2° S of E] makes South of East.

(I am studying for a test and I need to understand how to do this. Thanks in advanced)
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answers:
oubaas say: Determine the angle that the vector sum of the displacements Δd1 = 6.2 m [58.0° S of W], Δd2 = 6.8 m [S] and Δd3 = 28.0 m [17.2° S of E] makes South of East.

along x axis :
Δx = -6.2*cos 58 + 0 + 28*cos 17.2 = 23.46 m
along y axis :
Δy = -6.2*sin 58 - 6.8 - 28*sin 17.2 = -19.74 m

Δ = √23,46^2+(-19,74^2) = 30.66 m
heading = arctan Δy/Δx = -40.0° ( 40.0 SoE)
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electron1 say: To solve this problem, I suggest that east be the positive x direction.

For ∆d1,
South component = 6.2 * sin 58
West component = -6.2 * cos 58

For ∆3,
South component = 28 * sin 17.2
East = 28 * cos 17.2

Total south = 6.2 * sin 58 + 6.8 + 28 * sin 17.2
This is approximately 20 meters.

East = 28 * cos 17.2 – 6.2 * cos 58

This is approximately 23 meters. To determine the south of east, use the following equation.

Tan θ = South ÷ East

Tan θ = (6.2 * sin 58 + 6.8 + 28 * sin 17.2) ÷ (28 * cos 17.2 – 6.2 * cos 58)

The angle is approximately 40.9˚ south of east.

Displacement = √(South^2 + East^2)
Displacement = √[(6.2 * sin 58 + 6.8 + 28 * sin 17.2)^2 + (28 * cos 17.2 – 6.2 * cos 58)^2]
This is approximately 31 meters. I hope this is helpful for you.
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oldschool say: Express the angels on the 360° circle and sum the horizontal components, the vertical components and then d² = dx²+dy²

Sum horizontal components:
6.2*cos238 + 6.8*cos270 + 28*cos342.8 = 23.5
Sum vertical components:
6.2*sin238 + 6.8*sin270 + 28*sin342.8 = -20.3
Θ = arctan(-20.3/23.5) = -40.9° = 319.1° <<<<<
d² = 23.5²+20.3² = 31² <<<<<<
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