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.

Question

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

Answer 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