Question
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?
What was the length of the third leg of his trip?
Answers
bobpursley
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
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