A hiker makes three straight-line walks A 28 km at 289 degree B 31 km at 347 deg

Question

A hiker makes three straight-line walks A 28 km at 289 degree B 31 km at 347 degree C 18 km at 17 degree starting at position (41 km, 41 km), as illustrated. Which vector will return the hiker to the starting point? All angles are measured in a counter-clockwise direction from the positive x-axis. 1. ||D|| = 37.8851, theta_d = 227.664 2. ||D|| = 63.1712, km, theta_d = 153.502 degree 3. ||D|| = 28.7554, theta_d = 217.907 4. ||D|| = 24.4047, theta_d = 215.665 5. ||D|| = 43.5313, theta_d = 120.727 6. ||D|| = 10.2432, theta_d = 169.596 7. ||D|| = 23.4635, theta_d = 281.321 8. ||D|| = 13.4284, theta_d = 87.5432 9. ||D|| = 46.1045, theta_d = 135.868 10. ||D|| = 23.8904, theta_d = 298.681

Explanation / Answer

Given vectors A,B,C are
A 128 km ,at 289 degrees
B 31 km ,at 347 degrees
C 18 km ,at 17 degrees

we can resolve the vectors in to their components
A = Ax i + Ay j

A = A cos theta i + A sin theta j

A = 28 cos289 i + 28 sin289 j

A = 9.12 i - 26.47 j

for vector B

B = Bx i + By j

B= B cos theta i + B sin theta j

B = 31 cos347 i + 31 sin347 j

B = 30.20 i - 6.97 j

for vector C

C = Cx i + Cy j

C= C cos theta i + C sin theta j

C = 18cos17 i + 18 sin17 j

C = 17.21 i + 5.26 j

D = Dx i + Dy j

D = (Ax+Bx+Cx)i + (Ay+By+Cy) j

D = (9.12+30.20+17.21) i +(-26.47-6.97+5.26)j

D = 56.53 km i - 28.18km j

the magnitude is D = 63.17 km
the angle theta is = 26.5 degrees that is 180 -26.5 = 153.5 degrees so that the hiker will return to the starting point

option 2 is answer

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