Asked by Ms Fatimah
Two explorers leave a camp at the same time. one walks at 5km/h on a bearing 039°. the other walks at 7.5km/h on a bearing 265°. after two hours how far are they and what is the bearing of the second from the first.?
Answers
Answered by
oobleck
Use the law of cosines to find the distance z:
z^2 = 10^2 + 15^2 - 2*10*15 cos134°
To find the bearing, note that the two explorers are at positions
P=(10 sin39°,10cos39°) and Q=(-15cos5°,-15sin5°)
Thus, the bearing of #2 from #1 is 270-θ such that
tanθ = (P<sub><sub>y</sub></sub>-Q<sub><sub>y</sub></sub>)/(P<sub><sub>x</sub></sub>-Q<sub><sub>x</sub></sub>)
z^2 = 10^2 + 15^2 - 2*10*15 cos134°
To find the bearing, note that the two explorers are at positions
P=(10 sin39°,10cos39°) and Q=(-15cos5°,-15sin5°)
Thus, the bearing of #2 from #1 is 270-θ such that
tanθ = (P<sub><sub>y</sub></sub>-Q<sub><sub>y</sub></sub>)/(P<sub><sub>x</sub></sub>-Q<sub><sub>x</sub></sub>)
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