Question

Jason stands in the corner of a very large field. He walks, on a bearing of 030°, a distance of d metres. Jason then changes direction and walks twice as far on a new bearing of 120°. At the end of the walk Jason calculates both the distance he must
walk and the bearing required to return to his original position. Given that the total distance walked is 120 metres, what answers will Jason get if he is correct?​

Answer 1

Given:

Jason walks at 30° bearing of distance d meters and then from that point he walks at 120° bearing of distance 2d meters to a final position.

To Find:

We have to find the bearing and the distance that Jason has to walk to reach his final position to initial position.

Step-by-step explanation:

See the diagram attached first.

The triangle ΔOAB forms a right triangle with ∠OAB = 90° with distance OA = 40 meters and AB = 80 meters.

[Note: Since the total distance covered by Jason is given to be 120 meters and the distance AB = 2× distance OA, so AB = 80 meters and OA = 40 meters.]

Therefore, to return from final position of Jason i.e. B to initial position O, the bearing of walk will be 270° and the distance will be $= \sqrt{80^{2}+40^{2}} = 89.44 \textrm { meters}$