Physics

A pilot in a small plane encounters shifting winds. He flies 20.0 km at 30∘ north of east, then 50.0 km due north. From this point, he flies an additional distance in an unknown direction, only to find himself at a small airstrip that his map shows to be 90.0 km directly north of his starting point.

What was the length of the third leg of his trip?

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  1. first two legs:
    20cos30N+20sin30E + 50N
    third leg:
    xxx N + yyy W

    but you know the sum of all those is equal to 90N
    90N=N(20cos30+50+xxx)+E(20sin30+yyy)
    first, you know then xxx is 20sin30 W
    and yyy is 90-50-20cos30

    now, having the n and W components of the final leg, you can find distance with the pythoregean theorem

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    bobpursley

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